# 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

Click or tap a problem to see the solution.

### Example 1

Solve the equation

$4\cos^3x + 3\cos x - 7 = 0.$

### Example 2

Solve the trigonometric equation

$\cos x + \cos 7x - \cos 3x = 0.$

### Example 3

Solve the equation

$2\tan 3x + \sin 3x = 0.$

### Example 4

Solve the trigonometric equation

$\sin x + \sin 3x = 4\cos^3x.$

### 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.

$\sin x - \cos x = 0, \Rightarrow \tan x = 1, \Rightarrow x_2 = \arctan 1 + \pi k = \frac{\pi}{4} + \pi k, k \in \mathbb{Z}.$

$x_1 = \frac{\pi}{2} + \pi n; \;x_2 = \frac{\pi}{4} + \pi k; \;n, k \in \mathbb{Z}.$