Solving Trigonometric Equations by Factoring
In many cases, the left side of the trigonometric equation
\[f\left({x}\right) = 0\]
can be factored. If f (x) can be represented as
\[f\left({x}\right) = f_1\left({x}\right)f_2\left({x}\right)\cdots f_n\left({x}\right),\]
then the solution of the original equation will be all the roots of the equations
\[f_1\left({x}\right)= 0, f_2\left({x}\right) = 0, \ldots, f_n\left({x}\right) = 0,\]
which belong to the domain of the original equation f (x) = 0.
Factoring can be done in various ways: by the grouping method, using short multiplication formulas, etc.
For quadratic expressions, the following factoring formula is true:
\[ax^2 + bx +c = a\left({x - x_1}\right)\left({x - x_2}\right),\]
where \(x_1\) and \(x_2\) are the roots of the equation
\[ax^2 + bx + c = 0.\]
For example, the trigonometric equation
\[2\sin^2x - 3\sin x + 1 = 0\]
contains a quadratic expression whose roots are \({\frac{1}{2}}\) and \(1.\) In this case, this equation is represented in the form
\[2\left({\sin x - \frac{1}{2}}\right)\left({\sin x - 1}\right) = 0.\]
As you can see, the problem is reduced to solving two basic trigonometric equations:
\[\sin x = \frac{1}{2} \text{ and } \sin x = 1.\]
Solved Problems
Example 1.
Solve the equation
\[4\cos^3x + 3\cos x - 7 = 0.\]
Solution.
Note that \(\cos x = 1\) is a solution to the equation. Therefore, we can group and factor the left-hand side of the equation as follows:
\[4\cos^3x + 3\cos x - 7 = 0,\]
\[\Rightarrow 4\cos^3x \underbrace{\;-\;4\cos^2x + 4\cos^2x}_{0} \underbrace{\;-\;4\cos x +7\cos x}_{3\cos x} - 7 = 0,\]
\[\Rightarrow \left({4\cos^3x - 4\cos^2x}\right) + \left({4\cos^2x - 4\cos x}\right) + 7\cos x - 7 = 0,\]
\[\Rightarrow 4\cos^2x\left({\cos x - 1}\right) + 4\cos x\left({\cos x - 1}\right) +7\left({\cos x - 1}\right) = 0,\]
\[\Rightarrow \left({\cos x - 1}\right)\left({4\cos^2x +4\cos x + 7}\right) = 0.\]
Equating to zero each of the factors on the left side of the equation, we get:
Case 1.
\[\cos x = 1, \Rightarrow x = 2\pi n, n \in \mathbb{Z}.\]
Case 2.
\[4\cos^2x +4\cos x + 7 = 0.\]
Let \(t = \cos x.\) The quadratic equation \(4t^2 +4t + 7 = 0\) has no solutions since its discriminant is negative:
\[D = \left({-4}\right)^2 - 4\cdot 4\cdot 7 = -96 \lt 0.\]
Example 2.
Solve the trigonometric equation
\[\cos x + \cos 7x - \cos 3x = 0.\]
Solution.
Using the sum of cosines identity we can write the first two terms as
\[\cos x + \cos 7x = 2\cos\frac{x+7x}{2}\cos\frac{x-7x}{2} = 2\cos 4x \cos \left({-3x}\right) = 2\cos 4x \cos 3x.\]
Now we can factor the left side of the equation:
\[2\cos 4x \cos 3x - \cos 3x = 0, \Rightarrow \cos 3x\left({2\cos 4x - 1}\right) = 0.\]
We get two basic equations:
Case 1.
\[\cos 3x = 0, \Rightarrow 3x = \frac{\pi}{2} + \pi n, \Rightarrow x_1 = \frac{\pi}{6} + \frac{\pi n}{3}, n \in \mathbb{Z}.\]
Case 2.
\[2\cos 4x - 1, \Rightarrow \cos 4x = \frac{1}{2}, \Rightarrow 4x = \pm \arccos \frac{1}{2} + 2\pi k= \pm \frac{\pi}{3} + 2\pi k, \Rightarrow x_2 = \pm\frac{\pi}{12} + \frac{\pi k}{2}, k \in \mathbb{Z}.\]
So we have
\[x_1 = \frac{\pi}{6} + \frac{\pi n}{3};\; x_2 = \pm\frac{\pi}{12} + \frac{\pi k}{2};\; n,k \in \mathbb{Z}.\]
Example 3.
Solve the equation
\[2\tan 3x + \sin 3x = 0.\]
Solution.
It's obvious that
\[2\tan 3x + \sin 3x = 0, \Rightarrow 2\frac{\sin 3x}{\cos 3x} + \sin 3x = 0.\]
We can factor the left-hand side:
\[\sin 3x\left({\frac{2}{\cos 3x} + 1}\right) = 0.\]
Solve the two equations:
Case 1.
\[\sin 3x = 0, \Rightarrow 3x = \pi n, \Rightarrow x_1 = \frac{\pi n}{3}, n \in \mathbb{Z}.\]
Case 2.
\[\frac{2}{\cos 3x} + 1 = 0, \Rightarrow \cos 3x = -2, \Rightarrow x_2 \in \varnothing.\]
Make sure that the root \(x_1\) satisfies the domain of the equation:
\[\cos 3x \ne 0, \Rightarrow 3x \ne \frac{\pi}{2} + \pi n, \Rightarrow x \ne \frac{\pi}{6} + \frac{\pi n}{3}.\]
It can be seen that points \(x_1 = \frac{\pi n}{3}\) do not intersect with points \(x = \frac{\pi}{6} + \frac{\pi n}{3}.\)
Example 4.
Solve the trigonometric equation
\[\sin x + \sin 3x = 4\cos^3x.\]
Solution.
We transform the left side of the equation into product using the sum of sines formula:
\[2\sin\frac{x+3x}{2}\cos\frac{x-3x}{2} = 4\cos^3x, \Rightarrow 2\sin 2x \cos \left({-x}\right) = 4\cos^3x.\]
We take into account that the cosine function is even. Then we have
\[2\sin 2x \cos x - 4\cos^3x = 0.\]
By the double-angle identity \(\sin 2x = 2\sin x \cos x.\) This yields:
\[4\sin x \cos^2x -4\cos^3x = 0, \Rightarrow 4\cos^2x\left({\sin x - \cos x}\right) = 0.\]
The original equation was reduced to solving two elementary equations:
Case 1.
\[\cos^2x = 0, \Rightarrow \cos x = 0, \Rightarrow x_1 = \frac{\pi}{2} + \pi n, n \in \mathbb{Z}.\]
Case 2.
\[\sin x - \cos x = 0, \Rightarrow \tan x = 1, \Rightarrow x_2 = \arctan 1 + \pi k = \frac{\pi}{4} + \pi k, k \in \mathbb{Z}.\]
Answer:
\[x_1 = \frac{\pi}{2} + \pi n; \;x_2 = \frac{\pi}{4} + \pi k; \;n, k \in \mathbb{Z}.\]
