Symmetric Trigonometric Equations
Consider a rational expression of kind R(sin x, cos x) depending on the functions sin x and cos x. The expression R(sin x, cos x) is called symmetric if
\[R\left({\sin x,\cos x}\right) = R\left({\cos x,\sin x}\right).\]
For example, the following expressions are symmetric:
\[\sin^2x+3\sin x\cos x+\cos^2x-2,\]
\[2\sin^4x - \sin^2x \cos^2x + 2\cos^4x.\]
Symmetric trigonometric equations involving sin x and cos x can be solved using the substitution
\[z=\sin x+\cos x.\]
The product of sine and cosine is easily expressed in terms of their sum. Indeed:
\[z^2=\left({\sin x + \cos x}\right)^2=\sin^2x+2\sin x\cos x + \cos^2x=1+2\sin x\cos x,\]
\[\Rightarrow \sin x\cos x = \frac{z^2-1}{2}.\]
Example
Solve the equation:
\[\sin x+2\sin x \cos x +\cos x = 1.\]
Solution.
By swapping sin x and cos x we make sure that this equation is symmetric:
\[R\left({\sin x, \cos x}\right)=\sin x+2\sin x\cos x +\cos x-1=\cos x+2\cos x\sin x +\sin x-1=R\left({\cos x, \sin x}\right).\]
Introduce the new variable
\[z=\sin x+\cos x.\]
Then
\[\sin x\cos x = \frac{z^2-1}{2}.\]
Substituting this into the equation we rewrite it in the form
\[\sin x+2\cos x\sin x +\cos x-1=z+\frac{\cancel{2}\left({z^2-1}\right)}{\cancel{2}}-1=z^2+z-2=0.\]
We got a quadratic equation for z. Find its roots:
\[z_{1,2}=\frac{-1\pm\sqrt{1-4\cdot\left({-2}\right)}}{2}=\frac{-1\pm3}{2}=-2,1.\]
We have two roots and, respectively, two basic trigonometric equations:
\[\sin x + \cos x = -2, \;\;\sin x + \cos x = 1.\]
Using the cofunction identity and sum of sines formula, we can write
\[\sin x + \cos x = \sin x + \sin\left({\frac{\pi}{2}-x}\right)=2\sin\frac{\cancel{x}+\frac{\pi}{2}-\cancel{x}}{2}\cos\frac{x-\left({\frac{\pi}{2}-x}\right)}{2}=2\sin\frac{\pi}{4}\cos\left({x-\frac{\pi}{4}}\right)=2\cdot\frac{\sqrt{2}}{2}\cos\left({x-\frac{\pi}{4}}\right)=\sqrt{2}\cos\left({x-\frac{\pi}{4}}\right).\]
Find the solutions for x:
Case 1.
\[\sqrt{2}\cos\left({x-\frac{\pi}{4}}\right) = -2, \Rightarrow \cos\left({x-\frac{\pi}{4}}\right) = -\frac{2}{\sqrt{2}}=-\sqrt{2} \lt -1,\]
which means that \(x \in \varnothing.\)
Case 2.
\[\sqrt{2}\cos\left({x-\frac{\pi}{4}}\right) =1, \Rightarrow \cos\left({x-\frac{\pi}{4}}\right) =\frac{1}{\sqrt{2}}, \Rightarrow x-\frac{\pi}{4}=\pm\arccos\frac{1}{\sqrt{2}}+2\pi n=\pm\frac{\pi}{4} +2\pi n, \Rightarrow x = \frac{\pi}{4} \pm \frac{\pi}{4} +2\pi n,\]
or
\[x=\left[{\begin{array}{*{20}{l}}
{\frac{\pi}{2} + 2\pi n }\\
{2\pi k}
\end{array}}\right., \text{ where } n,k \in \mathbb{Z}.\]
In addition to sin x and cos x, symmetric equations may include other pairs of trigonometric functions, for example
\[\tan x \text { and } \cot x,\;\;\cos 2x \text { and } \frac{1}{\cos 2x},\;\;\sin \frac{x}{2} \text { and } -\cos \frac{x}{2},\]
and so on.
Some symmetric equations are considered in the examples below.
Solved Problems
Example 1.
Solve the equation
\[\sqrt{2}\left({\sin x + \cos x}\right) + \sin x \cos x =-\frac{1}{2}.\]
Solution.
This equation is symmetric since
\[R\left({\sin x, \cos x}\right) = \sqrt{2}\left({\sin x + \cos x}\right) + \sin x \cos x +\frac{1}{2}=\sqrt{2}\left({\cos x + \sin x}\right) + \cos x \sin x +\frac{1}{2}=R\left({\cos x, \sin x}\right).\]
Make the substitution
\[z = \sin x + \cos x.\]
In terms of the variable z, the equation takes the form
\[\sqrt{2}z+\frac{z^2-1}{2}=-\frac{1}{2}, \Rightarrow 2\sqrt{2}z+z^2-\cancel{1}=-\cancel{1}, \Rightarrow z^2+2\sqrt{2}z=0,\]
or
\[z\left({z+2\sqrt{2}}\right)=0.\]
The first value z1 = 0 produces the following solutions for x:
\[z_1=0,\Rightarrow \sin x +\cos x =0,\Rightarrow \tan x =-1, \Rightarrow x_1 =-\frac{\pi}{4} + \pi n,n \in \mathbb{Z}.\]
Consider the second value of z:
\[z_2+2\sqrt{2}=0,\Rightarrow \sin x + \cos x=-2\sqrt{2},\Rightarrow \sqrt{2}\cos\left({x-\frac{\pi}{4}}\right)=-2\sqrt{2},\Rightarrow \cos\left({x-\frac{\pi}{4}}\right)=-2 \lt -1,\]
that is, in this case \(x \in \varnothing.\) The final answer is given by
\[x=-\frac{\pi}{4}+\pi n,n\in \mathbb{Z}.\]
Example 2.
Solve the equation
\[\sin^3x+\cos^3x+\sin x\cos x=1.\]
Solution.
Obviously, this equation is symmetric:
\[R\left({\sin x,\cos x}\right)=\sin^3x+\cos^3x+\sin x\cos x-1=\cos^3x+\sin^3x+\cos x\sin x-1=R\left({\cos x,\sin x}\right).\]
Using the cube of sum formula, we have
\[\left({\sin x + \cos x}\right)^3=\sin^3x+3\sin^2x\cos x+3\sin x\cos^2x+\cos^3x=\sin^3x+3\sin x\cos x\left({\sin x + \cos x}\right) + \cos^3x.\]
Hence
\[\sin^3x+\cos^3x=\left({\sin x + \cos x}\right)^3 - 3\sin x\cos x\left({\sin x + \cos x}\right).\]
We introduce the new variable t:
\[t=\sin x + \cos x,\Rightarrow \sin x\cos x = \frac{t^2-1}{2}.\]
In terms of t, the equation is written as
\[t^3-\frac{3\left({t^2-1}\right)t}{2}+\frac{t^2-1}{2}-1=0,\Rightarrow t^3-\frac{3t^3}{2}+\frac{3t}{2}+\frac{t^2}{2}-\frac{1}{2}-1=0,\Rightarrow -\frac{t^3}{2}+\frac{3t}{2}+\frac{t^2}{2}-\frac{3}{2}=0,\Rightarrow t^3-3t-t^2+3=0.\]
We can factor the cubic equation:
\[t^3-3t-t^2+3=0,\Rightarrow t\left({t^2-3}\right)-\left({t^2-3}\right)=0,\Rightarrow\left({t^2-3}\right)\left({t-1}\right)=0.\]
This gives us the following possible solutions:
Case 1.
\[t^2-3=0,\Rightarrow \left({\sin x + \cos x}\right)^2=3,\Rightarrow \sin^2x+2\sin x\cos x + \cos^2x=3,\Rightarrow 1+\sin2x=3,\Rightarrow \sin2x=2>1,\]
so here we get \(x \in \varnothing.\)
Case 2.
\[t-1=0,\Rightarrow \sin x + \cos x =1,\Rightarrow\sqrt{2}\cos\left({x-\frac{\pi}{4}}\right)=1,\Rightarrow\cos\left({x-\frac{\pi}{4}}\right)=\frac{1}{\sqrt{2}},\Rightarrow x =\frac{\pi}{4}\pm\arccos\frac{1}{\sqrt{2}}+2\pi n=\frac{\pi}{4}\pm\frac{\pi}{4}+2\pi n,n\in\mathbb{Z}.\]
This solution can be represented in the form
\[x =\left[{\begin{array}{*{20}{l}}
{\frac{\pi}{2} + 2\pi n }\\
{2\pi k}
\end{array}}\right.,n,k \in \mathbb{Z}.\]
Example 3.
Solve the trigonometric equation
\[\sin^2x+\frac{1}{\sin^2x}=\sin x+\frac{1}{\sin x}.\]
Solution.
It is easy to see that this equation is symmetric with respect to the functions \(\sin x\) and \(\frac{1}{\sin x}.\) We denote
\[u = \sin x + \frac{1}{\sin x}.\]
Then
\[u^2=\left({\sin x + \frac{1}{\sin x}}\right)^2=\sin^2x+\frac{2\cancel{\sin x}}{\cancel{\sin x}}+\frac{1}{\sin^2x}=\sin^2x+2+\frac{1}{\sin^2x},\]
so
\[\sin^2x+\frac{1}{\sin^2x} = u^2 -2.\]
In terms of u, the equation takes the form
\[u^2-2=u, \Rightarrow u^2-u-2=0.\]
Solve the quadratic equation:
\[D = \left({-1}\right)^2-4\cdot \left({-2}\right)=9,\Rightarrow u_{1,2}=\frac{1\pm\sqrt{9}}{2}=\frac{1\pm 3}{2}=-1,2.\]
Hence, we have two cases.
Case 1.
\[\sin x+\frac{1}{\sin x }=-1, \Rightarrow \sin^2x+\sin x+1=0.\]
Calculate the discriminant:
\[D=1^2-4\cdot 1=-3\lt0.\]
Since D < 0, this equation does not have roots: \(x \in \varnothing.\)
Case 2.
\[\sin x+\frac{1}{\sin x }=2, \Rightarrow \sin^2x-2\sin x+1=0,\Rightarrow \left({\sin x - 1}\right)^2=0,\Rightarrow \sin x =1,\Rightarrow x = \frac{\pi}{2}+2\pi n,n\in\mathbb{Z}.
\]
Example 4.
Solve the trigonometric equation
\[3\left({\tan^2x+\cot^2x}\right)+4\left({\tan x+\cot x}\right)=-2.\]
Solution.
Obviously, this equation is symmetric with respect to tangent and cotangent. Let
\[z = \tan x + \cot x.\]
Notice that
\[z^2=\left({\tan x + \cot x}\right)^2=\tan^2x+2\,\underbrace{\tan x\cot x}_{1} +\cot^2x=\tan^2x+2 +\cot^2x.\]
Therefore
\[\tan^2x+\cot^2x=z^2-2.\]
Plugging this into the original equation gives
\[3\left({z^2-2}\right)+4z=-2,\Rightarrow 3z^2+4z-4=0.\]
Solve this quadratic equation:
\[D=4^2-4\cdot3\cdot\left({-4}\right)=64,\Rightarrow z_{1,2}=\frac{-4\pm 8}{6}=-2,\frac{2}{3}.\]
Consider these two cases and find solutions for x:
Case 1.
\[\tan x+\cot x=-2,\Rightarrow \tan^2x+2\tan x+1=0,\Rightarrow \left({\tan x + 1}\right)^2=0,\Rightarrow \tan x=-1,\Rightarrow x=-\frac{\pi}{4}+\pi n,n\in\mathbb{Z}.\]
Case 2.
\[\tan x+\cot x=\frac{2}{3},\Rightarrow \tan^2x-\frac{2}{3}\tan x+1=0,\Rightarrow 3\tan^2x-2\tan x+3=0.\]
We see that in the last equation
\[D=\left({-2}\right)^2-4\cdot3\cdot3=-32\lt0.\]
Accordingly, in this case \(x \in \varnothing.\)
