Nonlinear equations in two unknowns
Definition 1 . Let A be some set of pairs of numbers (x; y) . It is said that the set A is given numeric function z from two variables x and y , if a rule is specified, with the help of which a certain number is assigned to each pair of numbers from the set A.
Specifying a numerical function z of two variables x and y is often designate So:
Where f (x , y) - any function other than the function
f (x , y) = ax+by+c ,
where a, b, c are given numbers.
Definition 3 . Equation (2) solution name a pair of numbers x; y) , for which formula (2) is a true equality.
Example 1 . solve the equation
Since the square of any number is non-negative, it follows from formula (4) that the unknowns x and y satisfy the system of equations
the solution of which is a pair of numbers (6 ; 3) .
Answer: (6; 3)
Example 2 . solve the equation
Therefore, the solution to equation (6) is an infinite number of pairs of numbers kind
(1 + y ; y) ,
where y is any number.
linear
Definition 4 . Solving the system of equations
name a pair of numbers x; y) , substituting them into each of the equations of this system, we obtain the correct equality.
Systems of two equations, one of which is linear, have the form
g(x , y)
Example 4 . Solve a system of equations
Solution . Let us express the unknown y from the first equation of system (7) in terms of the unknown x and substitute the resulting expression into the second equation of the system:
Solving the Equation
x 1 = - 1 , x 2 = 9 .
Hence,
y 1 = 8 - x 1 = 9 ,
y 2 = 8 - x 2 = - 1 .
Systems of two equations, one of which is homogeneous
Systems of two equations, one of which is homogeneous, have the form
where a , b , c are given numbers, and g(x , y) is a function of two variables x and y .
Example 6 . Solve a system of equations
Solution . Let's solve the homogeneous equation
3x 2 + 2xy - y 2 = 0 ,
3x 2 + 17xy + 10y 2 = 0 ,
treating it as a quadratic equation with respect to the unknown x:
.
In case when x = - 5y, from the second equation of system (11) we obtain the equation
5y 2 = - 20 ,
which has no roots.
In case when
from the second equation of system (11) we obtain the equation
,
whose roots are numbers y 1 = 3 , y 2 = - 3 . Finding for each of these values y the corresponding value x , we obtain two solutions to the system: (- 2 ; 3) , (2 ; - 3) .
Answer: (- 2 ; 3) , (2 ; - 3)
Examples of solving systems of equations of other types
Example 8 . Solve the system of equations (MIPT)
Solution . We introduce new unknowns u and v , which are expressed in terms of x and y by the formulas:
To rewrite system (12) in terms of new unknowns, we first express the unknowns x and y in terms of u and v . It follows from system (13) that
We solve the linear system (14) by excluding the variable x from the second equation of this system. To this end, we perform the following transformations on system (14):
- we leave the first equation of the system unchanged;
- subtract the first equation from the second equation and replace the second equation of the system with the resulting difference.
As a result, system (14) is transformed into an equivalent system
from which we find
Using formulas (13) and (15), we rewrite the original system (12) as
The first equation of system (16) is linear, so we can express the unknown u from it in terms of the unknown v and substitute this expression into the second equation of the system.
In this article, we will consider a method for solving homogeneous trigonometric equations.
Homogeneous trigonometric equations have the same structure as homogeneous equations of any other kind. Let me remind you how to solve homogeneous equations of the second degree:
Consider homogeneous equations of the form
Distinctive features of homogeneous equations:
a) all monomials have the same degree,
b) the free term is equal to zero,
c) the equation contains powers with two different bases.
Homogeneous equations are solved by a similar algorithm.
To solve this type of equation, divide both sides of the equation by (can be divided by or by )
Attention! When dividing the right and left sides of the equation by an expression containing an unknown, you can lose the roots. Therefore, it is necessary to check whether the roots of the expression by which we divide both parts of the equation are the roots of the original equation.
If it is, then we write out this root so that we don’t forget about it later, and then we divide by this expression.
In general, the first thing to do when solving any equation with a zero on the right side is to try to factorize the left side of the equation in any way possible. And then set each factor to zero. In this case, we definitely will not lose the roots.
So, carefully divide the left side of the equation into an expression term by term. We get:
Reduce the numerator and denominator of the second and third fractions:
Let's introduce a replacement:
We get a quadratic equation:
We solve the quadratic equation, find the values, and then return to the original unknown.
When solving homogeneous trigonometric equations, there are a few important things to remember:
1. The free term can be converted to the square of sine and cosine using the basic trigonometric identity:
2. The sine and cosine of a double argument are monomials of the second degree - the sine of a double argument can be easily converted to the product of sine and cosine, and the cosine of a double argument to the square of a sine or cosine:
Consider several examples of solving homogeneous trigonometric equations.
1 . Let's solve the equation:
This is a classic example of a homogeneous trigonometric equation of the first degree: the degree of each monomial is equal to one, the free term is equal to zero.
Before dividing both sides of the equation by , it is necessary to check that the roots of the equation are not the roots of the original equation. Check: if , then title="sin(x)0">, следовательно их сумма не равна нулю.!}
Divide both sides of the equation by .
We get: 
, Where
, Where
Answer: , Where
2. Let's solve the equation:
This is an example of a homogeneous trigonometric equation of the second degree. We remember that if we can factorize the left side of the equation, then it is desirable to do so. In this equation, we can take out the brackets. Let's do it:
Solution of the first equation: , where
The second equation is a homogeneous trigonometric equation of the first degree. To solve it, we divide both sides of the equation by . We get:
Answer: where
3 . Let's solve the equation:
To make this equation "become" homogeneous, we transform it into a product, and represent the number 3 as the sum of the squares of the sine and cosine:
We move all the terms to the left, open the brackets and give like terms. We get:
Let us factorize the left side and equate each factor to zero:
Answer: where
4 . Let's solve the equation:
We see what we can bracket. Let's do it:
Set each factor equal to zero:
Solution of the first equation:
The second set equation is a classical homogeneous equation of the second degree. The roots of the equation are not the roots of the original equation, so we divide both sides of the equation by:
Solution of the first equation:
Solution of the second equation.
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Today we will deal with homogeneous trigonometric equations. First, let's deal with the terminology: what is a homogeneous trigonometric equation. It has the following characteristics:
- it should have several terms;
- all terms must have the same degree;
- all functions included in a homogeneous trigonometric identity must necessarily have the same argument.
Solution algorithm
Separate the terms
And if everything is clear with the first point, then it is worth talking about the second in more detail. What does the same degree of terms mean? Let's look at the first task:
3cosx+5sinx=0
3\cos x+5\sin x=0
The first term in this equation is 3cosx 3\cos x. Note that there is only one trigonometric function here - cosx\cos x - and no other trigonometric functions are present here, so the degree of this term is 1. The same with the second - 5sinx 5 \ sin x - only the sine is present here, i.e. the degree of this term is also equal to one. So, we have before us an identity consisting of two elements, each of which contains a trigonometric function, and at the same time only one. This is a first degree equation.
Let's move on to the second expression:
4sin2 x+sin2x−3=0
4((\sin )^(2))x+\sin 2x-3=0
The first term of this construction is 4sin2 x 4((\sin )^(2))x.
Now we can write the following solution:
sin2 x=sinx⋅sinx
((\sin )^(2))x=\sin x\cdot \sin x
In other words, the first term contains two trigonometric functions, that is, its degree is two. Let's deal with the second element - sin2x\sin 2x. Recall the following formula - the double angle formula:
sin2x=2sinx⋅cosx
\sin 2x=2\sin x\cdot \cos x
And again, in the resulting formula, we have two trigonometric functions - sine and cosine. Thus, the power value of this member of the construction is also equal to two.
We turn to the third element - 3. From the high school mathematics course, we remember that any number can be multiplied by 1, so we write:
˜ 3=3⋅1
And the unit using the basic trigonometric identity can be written in the following form:
1=sin2 x⋅ cos2 x
1=((\sin )^(2))x\cdot ((\cos )^(2))x
Therefore, we can rewrite 3 as follows:
3=3(sin2 x⋅ cos2 x)=3sin2 x+3 cos2 x
3=3\left(((\sin )^(2))x\cdot ((\cos )^(2))x \right)=3((\sin )^(2))x+3(( \cos )^(2))x
Thus, our term 3 has been split into two elements, each of which is homogeneous and has a second degree. The sine in the first term occurs twice, the cosine in the second also occurs twice. Thus, 3 can also be represented as a term with an exponent of two.
Same with the third expression:
sin3 x+ sin2 xcosx=2 cos3 x
Let's get a look. The first term - sin3 x((\sin )^(3))x is a trigonometric function of the third degree. The second element is sin2 xcosx((\sin )^(2))x\cos x.
sin2 ((\sin )^(2)) is a link with a power value of two multiplied by cosx\cos x is the term of the first. In total, the third term also has a power value of three. Finally, on the right is another link - 2cos3 x 2((\cos )^(3))x is an element of the third degree. Thus, we have a homogeneous trigonometric equation of the third degree.
We have recorded three identities of different degrees. Notice again the second expression. In the original entry, one of the members has an argument 2x 2x. We are forced to get rid of this argument by transforming it according to the formula of the sine of a double angle, because all the functions included in our identity must necessarily have the same argument. And this is a requirement for homogeneous trigonometric equations.
We use the formula of the main trigonometric identity and write down the final solution
We figured out the terms, move on to the solution. Regardless of the power exponent, solving equalities of this type is always performed in two steps:
1) prove that
cosx≠0
\cos x\ne 0. To do this, it suffices to recall the formula for the basic trigonometric identity (sin2 x⋅ cos2 x=1)\left(((\sin )^(2))x\cdot ((\cos )^(2))x=1 \right) and substitute into this formula cosx=0\cosx=0. We will get the following expression:
sin2 x=1sinx=±1
\begin(align)& ((\sin )^(2))x=1 \\& \sin x=\pm 1 \\\end(align)
Substituting the obtained values, i.e. instead of cosx\cos x is zero, and instead of sinx\sin x - 1 or -1, in the original expression, we get an incorrect numerical equality. This is the rationale for the fact that
cosx≠0
2) the second step follows logically from the first. Because the
cosx≠0
\cos x\ne 0, we divide both sides of our construction by cosn x((\cos )^(n))x, where n n is the power exponent of the homogeneous trigonometric equation. What does this give us:
\[\begin(array)((35)(l))
sinxcosx=tgxcosxcosx=1
\begin(align)& \frac(\sin x)(\cos x)=tgx \\& \frac(\cos x)(\cos x)=1 \\\end(align) \\() \\ \end(array)\]
Due to this, our cumbersome initial construction reduces to the equation n n-power with respect to the tangent, the solution of which is easily written using a change of variable. That's the whole algorithm. Let's see how it works in practice.
We solve real problems
Task #1
3cosx+5sinx=0
3\cos x+5\sin x=0
We have already found out that this is a homogeneous trigonometric equation with a power exponent equal to one. Therefore, first of all, let us find out that cosx≠0\cos x\ne 0. Assume to the contrary that
cosx=0→sinx=±1
\cos x=0\to \sin x=\pm 1.
We substitute the resulting value into our expression, we get:
3⋅0+5⋅(±1)=0±5=0
\begin(align)& 3\cdot 0+5\cdot \left(\pm 1 \right)=0 \\& \pm 5=0 \\\end(align)
Based on this, it can be said that cosx≠0\cos x\ne 0. Divide our equation by cosx\cos x because our entire expression has a power value of one. We get:
3(cosxcosx) +5(sinxcosx) =0 3+5tgx=0tgx=− 3 5
\begin(align)& 3\left(\frac(\cos x)(\cos x) \right)+5\left(\frac(\sin x)(\cos x) \right)=0 \\& 3+5tgx=0 \\& tgx=-\frac(3)(5) \\\end(align)
This is not a table value, so the answer will include arctgx arctgx:
x=arctg (−3 5 ) + πn,n∈Z
x=arctg\left(-\frac(3)(5) \right)+\text( )\!\!\pi\!\!\text( )n,n\in Z
Because the arctg arctg arctg is an odd function, we can take the “minus” out of the argument and put it before arctg. We get the final answer:
x=−arctg 3 5 + πn,n∈Z
x=-arctg\frac(3)(5)+\text( )\!\!\pi\!\!\text( )n,n\in Z
Task #2
4sin2 x+sin2x−3=0
4((\sin )^(2))x+\sin 2x-3=0
As you remember, before proceeding with its solution, you need to perform some transformations. We perform transformations:
4sin2 x+2sinxcosx−3 (sin2 x+ cos2 x)=0 4sin2 x+2sinxcosx−3 sin2 x−3 cos2 x=0sin2 x+2sinxcosx−3 cos2 x=0
\begin(align)& 4((\sin )^(2))x+2\sin x\cos x-3\left(((\sin )^(2))x+((\cos )^(2 ))x \right)=0 \\& 4((\sin )^(2))x+2\sin x\cos x-3((\sin )^(2))x-3((\cos )^(2))x=0 \\& ((\sin )^(2))x+2\sin x\cos x-3((\cos )^(2))x=0 \\\end (align)
We have received a structure consisting of three elements. In the first term we see sin2 ((\sin )^(2)), i.e. its power value is two. In the second term, we see sinx\sin x and cosx\cos x - again, there are two functions, they are multiplied, so the total degree is again two. In the third link we see cos2 x((\cos )^(2))x - similar to the first value.
Let's prove that cosx=0\cos x=0 is not a solution to this construction. To do this, suppose the opposite:
\[\begin(array)((35)(l))
\cos x=0 \\\sin x=\pm 1 \\1+2\cdot \left(\pm 1 \right)\cdot 0-3\cdot 0=0 \\1+0-0=0 \ \1=0 \\\end(array)\]
We have proven that cosx=0\cos x=0 cannot be a solution. We pass to the second step - we divide our entire expression by cos2 x((\cos )^(2))x. Why in a square? Because the exponent of this homogeneous equation is equal to two:
sin2 xcos2 x+2sinxcosxcos2 x−3=0 t g2 x+2tgx−3=0
\begin(align)& \frac(((\sin )^(2))x)(((\cos )^(2))x)+2\frac(\sin x\cos x)(((\ cos )^(2))x)-3=0 \\& t((g)^(2))x+2tgx-3=0 \\\end(align)
Can this expression be solved using the discriminant? Of course you can. But I propose to recall the theorem converse to Vieta's theorem, and we get that this polynomial can be represented as two simple polynomials, namely:
(tgx+3) (tgx−1)=0tgx=−3→x=−arctg3+ π n,n∈Ztgx=1→x= π 4 + πk,k∈Z
\begin(align)& \left(tgx+3 \right)\left(tgx-1 \right)=0 \\& tgx=-3\to x=-arctg3+\text( )\!\!\pi\ !\!\text( )n,n\in Z \\& tgx=1\to x=\frac(\text( )\!\!\pi\!\!\text( ))(4)+\ text( )\!\!\pi\!\!\text( )k,k\in Z \\\end(align)
Many students ask whether it is worth writing separate coefficients for each group of solutions to identities, or not to bother and write the same coefficient everywhere. Personally, I think that it is better and more reliable to use different letters, so that in the case when you enter a serious technical university with additional tests in mathematics, the inspectors do not find fault with the answer.
Task #3
sin3 x+ sin2 xcosx=2 cos3 x
((\sin )^(3))x+((\sin )^(2))x\cos x=2((\cos )^(3))x
We already know that this is a homogeneous trigonometric equation of the third degree, no special formulas are needed, and all that is required of us is to transfer the term 2cos3 x 2((\cos )^(3))x to the left. Rewriting:
sin3 x+ sin2 xcosx−2 cos3 x=0
((\sin )^(3))x+((\sin )^(2))x\cos x-2((\cos )^(3))x=0
We see that each element contains three trigonometric functions, so this equation has a power value of three. We solve it. First of all, we need to prove that cosx=0\cos x=0 is not a root:
\[\begin(array)((35)(l))
\cos x=0 \\\sin x=\pm 1 \\\end(array)\]
Substitute these numbers into our original construction:
(±1)3 +1⋅0−2⋅0=0 ±1+0−0=0±1=0
\begin(align)& ((\left(\pm 1 \right))^(3))+1\cdot 0-2\cdot 0=0 \\& \pm 1+0-0=0 \\& \pm 1=0 \\\end(align)
Hence, cosx=0\cos x=0 is not a solution. We have proven that cosx≠0\cos x\ne 0. Now that we have proved this, we divide our original equation by cos3 x((\cos )^(3))x. Why in a cube? Because we just proved that our original equation has a third power:
sin3 xcos3 x+sin2 xcosxcos3 x−2=0 t g3 x+t g2 x−2=0
\begin(align)& \frac(((\sin )^(3))x)(((\cos )^(3))x)+\frac(((\sin )^(2))x\ cos x)(((\cos )^(3))x)-2=0 \\& t((g)^(3))x+t((g)^(2))x-2=0 \\\end(align)
Let's introduce a new variable:
tgx=t
Rewriting the structure:
t3 +t2 −2=0
((t)^(3))+((t)^(2))-2=0
We have a cubic equation. How to solve it? Initially, when I was just compiling this video tutorial, I planned to first talk about the decomposition of polynomials into factors and other tricks. But in this case, everything is much simpler. Look, our reduced identity, with the term with the highest degree, is 1. In addition, all the coefficients are integers. And this means that we can use the corollary of Bezout's theorem, which says that all roots are divisors of the number -2, that is, a free term.
The question arises: what is divided by -2. Since 2 is a prime number, there are not so many options. It can be the following numbers: 1; 2; -1; -2. Negative roots immediately disappear. Why? Because both of them are greater than 0 in absolute value, therefore, t3 ((t)^(3)) will be greater in modulus than t2 ((t)^(2)). And since the cube is an odd function, so the number in the cube will be negative, and t2 ((t)^(2)) is positive, and this whole construction, with t=−1 t=-1 and t=−2 t=-2 will not be greater than 0. Subtract -2 from it and get a number that is obviously less than 0. Only 1 and 2 remain. Let's substitute each of these numbers:
˜ t=1→ 1+1−2=0→0=0
˜t=1\to \text( )1+1-2=0\to 0=0
We got the correct numerical equality. Hence, t=1 t=1 is the root.
t=2→8+4−2=0→10≠0
t=2\to 8+4-2=0\to 10\ne 0
t=2 t=2 is not a root.
According to the corollary and the same Bezout theorem, any polynomial whose root is x0 ((x)_(0)), represent as:
Q(x)=(x= x0 )P(x)
Q(x)=(x=((x)_(0)))P(x)
In our case, as x x is a variable t t, and in the role x0 ((x)_(0)) is a root equal to 1. We get:
t3 +t2 −2=(t−1)⋅P(t)
((t)^(3))+((t)^(2))-2=(t-1)\cdot P(t)
How to find a polynomial P (t) P\left(t\right)? Obviously, you need to do the following:
P(t)= t3 +t2 −2 t−1
P(t)=\frac(((t)^(3))+((t)^(2))-2)(t-1)
We substitute:
t3 +t2 +0⋅t−2t−1=t2 +2t+2
\frac(((t)^(3))+((t)^(2))+0\cdot t-2)(t-1)=((t)^(2))+2t+2
So, our original polynomial is divided without a remainder. Thus, we can rewrite our original equality as:
(t−1)( t2 +2t+2)=0
(t-1)(((t)^(2))+2t+2)=0
The product is equal to zero when at least one of the factors is equal to zero. We have already considered the first factor. Let's look at the second one:
t2 +2t+2=0
((t)^(2))+2t+2=0
Experienced students probably already understood that this construction has no roots, but let's still calculate the discriminant.
D=4−4⋅2=4−8=−4
D=4-4\cdot 2=4-8=-4
The discriminant is less than 0, so the expression has no roots. In total, the huge construction was reduced to the usual equality:
\[\begin(array)((35)(l))
t=\text( )1 \\tgx=\text( )1 \\x=\frac(\text( )\!\!\pi\!\!\text( ))(4)+\text( ) \!\!\pi\!\!\text( )k,k\in Z \\\end(array)\]
In conclusion, I would like to add a couple of comments on the last task:
- whether the condition will always be satisfied cosx≠0\cos x\ne 0, and whether this check should be done at all. Of course, not always. In cases where cosx=0\cos x=0 is a solution to our equality, we should take it out of brackets, and then a full-fledged homogeneous equation will remain in brackets.
- What is the division of a polynomial by a polynomial. Indeed, most schools do not study this, and when students first see such a structure, they experience a slight shock. But, in fact, this is a simple and beautiful technique that greatly facilitates the solution of equations of higher degrees. Of course, a separate video tutorial will be devoted to it, which I will publish in the near future.
Key points
Homogeneous trigonometric equations are a favorite topic in various tests. They are solved very simply - it is enough to practice once. To make it clear what we are talking about, we introduce a new definition.
A homogeneous trigonometric equation is one in which each non-zero term of which consists of the same number of trigonometric factors. These can be sines, cosines, or combinations thereof - the solution method is always the same.
The degree of a homogeneous trigonometric equation is the number of trigonometric factors included in non-zero terms. Examples:
sinx+15 cos x=0
\sin x+15\text( cos )x=0 — 1st degree identity;
2 sin2x+5sinxcosx−8cos2x=0
2\text( sin)2x+5\sin xcosx-8\cos 2x=0 - 2nd degree;
sin3x+2sinxcos2x=0
\sin 3x+2\sin x\cos 2x=0 - 3rd degree;
sinx+cosx=1
\sin x+\cos x=1 - and this equation is not homogeneous, since there is a unit on the right - a non-zero term, in which there are no trigonometric factors;
sin2x+2sinx−3=0
\sin 2x+2\sin x-3=0 is also an inhomogeneous equation. Element sin2x\sin 2x - the second degree (because you can imagine
sin2x=2sinxcosx
\sin 2x=2\sin x\cos x), 2sinx 2 \ sin x - the first, and the term 3 is generally zero, since there are no sines or cosines in it.
General solution scheme
The solution scheme is always the same:
Let's pretend that cosx=0\cosx=0. Then sinx=±1\sin x=\pm 1 - this follows from the main identity. Substitute sinx\sin x and cosx\cos x into the original expression, and if the result is nonsense (for example, the expression 5=0 5=0), go to the second point;
We divide everything by the power of the cosine: cosx, cos2x, cos3x ... - depends on the power value of the equation. We get the usual equality with tangents, which is successfully solved after the replacement tgx=t.
tgx=tThe found roots will be the answer to the original expression.
With the help of this video lesson, students will be able to study the topic of homogeneous trigonometric equations.
Let's give definitions:
1) a homogeneous trigonometric equation of the first degree looks like a sin x + b cos x = 0;
2) a homogeneous trigonometric equation of the second degree looks like a sin 2 x + b sin x cos x + c cos 2 x = 0.
Consider the equation a sin x + b cos x = 0. If a is zero, then the equation will look like b cos x = 0; if b is zero, then the equation will look like a sin x = 0. These are the equations that we called the simplest and solved earlier in previous topics.
Now consider the option when a and b are not equal to zero. By dividing the parts of the equation by the cosine x and we will carry out the transformation. We get a tg x + b = 0, then tg x will be equal to - b / a.
From the above it follows that the equation a sin mx + b cos mx = 0 is a homogeneous trigonometric equation of the first degree. To solve an equation, divide its parts by cos mx.
Let's analyze example 1. Solve 7 sin (x / 2) - 5 cos (x / 2) = 0. First, divide the parts of the equation by cosine (x / 2). Knowing that the sine divided by the cosine is the tangent, we get 7 tg (x / 2) - 5 = 0. Transforming the expression, we find that the value of the tangent (x / 2) is 5/7. The solution to this equation is x = arctan a + πn, in our case x = 2 arctan (5/7) + 2πn.
Consider the equation a sin 2 x + b sin x cos x + c cos 2 x = 0:
1) with a equal to zero, the equation will look like b sin x cos x + c cos 2 x = 0. Transforming, we obtain the expression cos x (b sin x + c cos x) = 0 and proceed to the solution of two equations. After dividing the parts of the equation by the cosine x, we get b tg x + c = 0, which means tg x = - c/b. Knowing that x \u003d arctan a + πn, then the solution in this case will be x \u003d arctg (- c / b) + πn.
2) if a is not equal to zero, then, by dividing the parts of the equation by the cosine squared, we obtain an equation containing a tangent, which will be square. This equation can be solved by introducing a new variable.
3) when c equals zero, the equation will take the form a sin 2 x + b sin x cos x = 0. This equation can be solved by taking the sine of x out of the bracket.
1. see if there is a sin 2 x in the equation;
2. if the term a sin 2 x is contained in the equation, then the equation can be solved by dividing both parts by the cosine squared and then introducing a new variable.
3. if the equation a sin 2 x does not contain, then the equation can be solved by taking out the brackets cosx.
Consider example 2. We take out the cosine and get two equations. The root of the first equation is x = π/2 + πn. To solve the second equation, we divide the parts of this equation by the cosine x, by means of transformations we obtain x = π/3 + πn. Answer: x = π/2 + πn and x = π/3 + πn.
Let's solve example 3, an equation of the form 3 sin 2 2x - 2 sin 2x cos 2x + 3 cos 2 2x = 2 and find its roots that belong to the segment from - π to π. Because Since this equation is non-homogeneous, it is necessary to reduce it to a homogeneous form. Using the formula sin 2 x + cos 2 x = 1, we get the equation sin 2 2x - 2 sin 2x cos 2x + cos 2 2x = 0. Dividing all parts of the equation by cos 2 x, we get tg 2 2x + 2tg 2x + 1 = 0 Using the introduction of a new variable z = tg 2x, we solve the equation whose root is z = 1. Then tg 2x = 1, which implies that x = π/8 + (πn)/2. Because according to the condition of the problem, you need to find the roots that belong to the segment from - π to π, the solution will look like - π< x <π. Подставляя найденное значение x в данное выражение и преобразовывая его, получим - 2,25 < n < 1,75. Т.к. n - это целые числа, то решению уравнения удовлетворяют значения n: - 2; - 1; 0; 1. При этих значениях n получим корни решения исходного уравнения: x = (- 7π)/8, x = (- 3π)/8, x =π/8, x = 5π/8.
TEXT INTERPRETATION:
Homogeneous trigonometric equations
Today we will analyze how the "Homogeneous Trigonometric Equations" are solved. These are equations of a special kind.
Let's get acquainted with the definition.
Type equation and sinx+bcosx = 0 (and sine x plus be cosine x is zero) is called a homogeneous trigonometric equation of the first degree;
equation of the form a sin 2 x+bsin xcosx+ccos 2 x= 0 (and sine square x plus be sine x cosine x plus se cosine square x equals zero) is called a homogeneous trigonometric equation of the second degree.
If a=0, then the equation will take the form bcosx = 0.
If b = 0 , then we get and sin x = 0.
These equations are elementary trigonometric, and we considered their solution in our previous topics
Consider the case when both coefficients are non-zero. Divide both sides of the equation Asinx+ bcosx = 0 term by term on cosx.
We can do this, since cosine x is non-zero. After all, if cosx = 0 , then the equation Asinx+ bcosx = 0 will take the form Asinx = 0 , A≠ 0, therefore sinx = 0 . Which is impossible, because according to the basic trigonometric identity sin 2x+cos 2 x=1 .
Dividing both sides of the equation Asinx+ bcosx = 0 term by term on cosx, we get: + =0
Let's make the transformations:
1. Since = tg x, then =and tg x
2 reduce by cosx, Then
Thus we get the following expression and tg x + b =0.
Let's do the transformation:
1. Move b to the right side of the expression with the opposite sign
and tg x \u003d - b
2. Get rid of the multiplier and dividing both sides of the equation by a
tg x= -.
Conclusion: An equation of the form and sinmx+bcosmx = 0 (and the sine em x plus the cosine em x is zero) is also called a homogeneous trigonometric equation of the first degree. To solve it, divide both sides by cosmx.
EXAMPLE 1. Solve the equation 7 sin - 5 cos \u003d 0 (seven sine x by two minus five cosine x by two is zero)
Solution. We divide both parts of the equation term by term by cos, we get
1. \u003d 7 tg (since the ratio of sine to cosine is tangent, then seven sine x by two divided by cosine x by two is equal to 7 tangent x by two)
2. -5 = -5 (when abbreviated cos)
Thus we got the equation
7tg - 5 = 0, Let's transform the expression, move minus five to the right side, changing the sign.
We have reduced the equation to the form tg t = a, where t=, a =. And since this equation has a solution for any value A and these solutions look like
x \u003d arctg a + πn, then the solution to our equation will look like:
Arctg + πn, find x
x \u003d 2 arctan + 2πn.
Answer: x \u003d 2 arctg + 2πn.
Let's move on to a homogeneous trigonometric equation of the second degree
Asin 2 x+b sin x cos x +Withcos2 x= 0.
Let's consider several cases.
I. If a=0, then the equation will take the form bsinxcosx+ccos 2 x= 0.
When solving e then we use the factorization method of the equations. Let's take out cosx brackets and we get: cosx(bsinx+ccosx)= 0 . Where cosx= 0 or
b sin x +Withcos x= 0. And we already know how to solve these equations.
We divide both parts of the equation term by term by cosx, we get
1 (because the ratio of sine to cosine is the tangent).
Thus we get the equation: b tg x+c=0
We have reduced the equation to the form tg t = a, where t= x, a =. And since this equation has a solution for any value A and these solutions look like
x \u003d arctg a + πn, then the solution to our equation will be:
x \u003d arctg + πn, .
II. If a≠0, then we divide both parts of the equation term by term into cos 2 x.
(Arguing similarly, as in the case of a homogeneous trigonometric equation of the first degree, cosine x cannot vanish).
III. If c=0, then the equation will take the form Asin 2 x+ bsinxcosx= 0. This equation is solved by the factorization method (take out sinx for brackets).
So, when solving the equation Asin 2 x+ bsinxcosx+ccos 2 x= 0 you can follow the algorithm:
EXAMPLE 2. Solve the equation sinxcosx - cos 2 x= 0 (sine x times cosine x minus the root of three times cosine squared x equals zero).
Solution. Let us factorize (bracket cosx). Get
cos x(sin x - cos x)= 0, i.e. cos x=0 or sin x - cos x= 0.
Answer: x \u003d + πn, x \u003d + πn.
EXAMPLE 3. Solve the equation 3sin 2 2x - 2 sin2xcos2 x +3cos 2 2x= 2 (three sine square of two x minus twice the product of the sine of two x and the cosine of two x plus three cosine square of two x) and find its roots belonging to the interval (- π; π).
Solution. This equation is not homogeneous, so we will carry out transformations. The number 2 contained on the right side of the equation is replaced by the product 2 1
Since, according to the basic trigonometric identity, sin 2 x + cos 2 x \u003d 1, then
2 ∙ 1= 2 ∙ (sin 2 x + cos 2 x) = opening the brackets we get: 2 sin 2 x + 2 cos 2 x.
2 ∙ 1= 2 ∙ (sin 2 x + cos 2 x) =2 sin 2 x + 2 cos 2 x
So the equation 3sin 2 2x - 2 sin2xcos2 x +3cos 2 2x= 2 will take the form:
3sin 2 2x - 2sin 2x cos2 x +3cos 2 2x = 2 sin 2 x + 2 cos 2 x.
3sin 2 2x - 2 sin 2x cos2 x +3cos 2 2x - 2 sin 2 x - 2 cos 2 x=0,
sin 2 2x - 2 sin 2x cos2 x + cos 2 2x =0.
We have obtained a homogeneous trigonometric equation of the second degree. Let's apply the term-by-term division by cos 2 2x:
tg 2 2x - 2tg 2x + 1 = 0.
Let us introduce a new variable z= tg2x.
We have z 2 - 2 z + 1 = 0. This is a quadratic equation. Noticing the abbreviated multiplication formula on the left side - the square of the difference (), we get (z - 1) 2 = 0, i.e. z = 1. Let's return to the reverse substitution:
We have reduced the equation to the form tg t \u003d a, where t \u003d 2x, a \u003d 1. And since this equation has a solution for any value A and these solutions look like
x \u003d arctg x a + πn, then the solution to our equation will be:
2x \u003d arctg1 + πn,
x \u003d + , (x is equal to the sum of pi times eight and pi en times two).
It remains for us to find such values of x that are contained in the interval
(- π; π), i.e. satisfy the double inequality - π x π. Because
x= + , then - π + π. Divide all parts of this inequality by π and multiply by 8, we get
move the unit to the right and to the left, changing the sign to minus one
divide by four we get
for convenience, in fractions, we select integer parts
- This inequality is satisfied by the following integer n: -2, -1, 0, 1
