# Precalculus

## Trigonometry # Irrational Trigonometric Equations

An irrational trigonometric equation is an equation containing a trigonometric function under the sign of a root of any degree. We'll consider here equations of the form

$\sqrt{f\left({x}\right)} = g\left({x}\right),$

where $${f\left({x}\right)},$$ $${g\left({x}\right)}$$ are trig functions.

When solving an irrational equation, as a rule, we make transition to an equivalent system. In this case, the equivalent transition looks like this

$\sqrt{f\left({x}\right)} = g\left({x}\right) \Leftrightarrow \left\{ \begin{array}{l} f\left({x}\right) = g^2\left({x}\right)\\ g\left({x}\right) \ge 0 \end{array} \right..$

Note that we require the right-hand side of the equation $${g\left({x}\right)}$$ to be non-negative. Since $$g^2\left({x}\right) \ge 0,$$ it automatically turns out that $${f\left({x}\right)} \ge 0.$$

Consider examples.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Solve the equation

$\sqrt{2 + \sin x} = 2 + \sin x.$

### Example 2

Solve the trigonometric equation

$\sqrt{\tan x} + \sqrt{\cot x} = 2.$

### Example 3

Solve the trigonometric equation

$\sin x = \sqrt{1 - \cos x}.$

### Example 4

Solve the equation

$\sqrt{\sin x} + \cos x = 0.$

### Example 5

Solve the equation

$\sin2x + \sqrt{2 - 5\sin^2x} = 0.$

### Example 6

Solve the trigonometric equation

$4\sin x +\sqrt{6 - 6\tan^2x} = 0.$

### Example 1.

Solve the equation

$\sqrt{2 + \sin x} = 2 + \sin x.$

Solution.

Let's move on to the equivalent system:

$\left\{ \begin{array}{l} 2 + \sin x = \left({2 + \sin x}\right)^2\\ 2 + \sin x \ge 0 \end{array} \right..$

Let $$2 + \sin x = t.$$ Then we get the following simple system:

$\left\{ \begin{array}{l} t = t^2\\ t \ge 0 \end{array} \right..$

The roots of the equation are $$t_1 = 0, t_2 = 1.$$ Both roots satisfy the condition $$t \ge 0.$$

Back to variable $$x:$$

#### Solution 1.

$2 + \sin x = t_1 = 0, \Rightarrow \sin x = -2, \Rightarrow x = \varnothing.$

#### Solution 2.

$2 + \sin x = t_2 = 1, \Rightarrow \sin x = -1, \Rightarrow x = -\frac{\pi}{2} + 2\pi n,\;n\in\mathbb{Z}.$

The final answer is given by

$x = -\frac{\pi}{2} + 2\pi n,\;n\in\mathbb{Z}.$

### Example 2.

Solve the trigonometric equation

$\sqrt{\tan x} + \sqrt{\cot x} = 2.$

Solution.

We represent the equation in the following form:

$\sqrt{\tan x} + \sqrt{\cot x} = 2, \Rightarrow \sqrt{\tan x} + \sqrt{\frac{1}{\tan x}} = 2, \Rightarrow \sqrt{\tan x} + {\frac{1}{\sqrt{\tan x}}} = 2.$

Introduce the new variable $$t = \sqrt{\tan x}$$ assuming $$t \ne 0.$$ Then the equation becomes

$t + \frac{1}{t} = 2.$

Multiplying both sides by $$t$$ we find the solution of the equation:

$t^2 - 2t + 1 = 0, \Rightarrow \left({t - 1}\right)^2 = 0, \Rightarrow t - 1 = 0, \Rightarrow t = 1.$

So we get

$\sqrt{\tan x} = 1, \Rightarrow \tan x = 1, \Rightarrow x = \arctan{1} + \pi n = \frac{\pi}{4} + \pi n,\;n\in\mathbb{Z}.$

### Example 3.

Solve the trigonometric equation

$\sin x = \sqrt{1 - \cos x}.$

Solution.

We write an equivalent system for this equation:

$\left\{ \begin{array}{l} 1 - \cos x = \sin^2x\\ \sin x \ge 0 \end{array} \right..$

Solve the resulting equation:

$1 - \cos x = \sin^2x, \Rightarrow \cancel{1} - \cos x = \cancel{1} - \cos^2x, \Rightarrow \cos^2x - \cos x = 0, \Rightarrow \cos x \left({\cos x - 1}\right) = 0.$

We get two roots.

#### Solution 1.

$\cos x = 0, \Rightarrow x_1 = \frac{\pi}{2} + \pi n,\;n\in\mathbb{Z}.$

#### Solution 2.

$\cos x = 1, \Rightarrow x_2 = 2\pi k,\;k\in\mathbb{Z}.$

Mark these points on the unit circle.

Possible solutions are limited by inequality $$\sin x \ge 0$$ and must lie in the upper half-plane. The point $$x = -\frac{\pi}{2}$$ is not in this area. Therefore, the solution of the equation is described by two branches:

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

### Example 4.

Solve the equation

$\sqrt{\sin x} + \cos x = 0.$

Solution.

We transform this equation into an equivalent system:

$\sqrt{\sin x} = -\cos x \Leftrightarrow \left\{ \begin{array}{l} \sin x = \left({-\cos x}\right)^2\\ -\cos x \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \sin x = \cos^2x\\ \cos x \le 0 \end{array} \right..$

Solve the upper equation of the system:

$\sin x = \cos^2x, \Rightarrow \sin x = 1 - \sin^2x, \Rightarrow \sin^2x + \sin x - 1 = 0.$

Denote $$\sin x = t$$ and find the roots of the quadratic equation:

$t^2 + t - 1 = 0,\; D = 1^2 - 4\cdot\left({-1}\right) = 5, \Rightarrow t_{1,2} = \frac{-1 \pm \sqrt{5}}{2}.$

Note that the value of $$t_1$$ with a minus sign in front of the square root lies in the following range:

$2 \lt \sqrt{5} \lt 3,$
$\Rightarrow -3 \lt -\sqrt{5} \lt -2,$
$\Rightarrow -4 \lt -1-\sqrt{5} \lt -3,$
$\Rightarrow -2 \lt \frac{-1-\sqrt{5}}{2} \lt -\frac{3}{2}.$

Since $$-1 \le \sin x \le 1,$$ then $$x = \varnothing$$ for $$t_1 = \frac{-1-\sqrt{5}}{2}.$$

Estimate the value of $$t_2:$$

$2 \lt \sqrt{5} \lt 3,$
$\Rightarrow 1 \lt \sqrt{5} - 1 \lt 2,$
$\Rightarrow \frac{1}{2} \lt \frac{\sqrt{5} - 1}{2} \lt 1.$

Here the equation has the following solution:

$\sin x = t_2 = \frac{\sqrt{5} - 1}{2}, \Rightarrow x = \left({-1}\right)^n \arcsin{\frac{\sqrt{5} - 1}{2}} + \pi n,$

or

$\left[ \begin{array}{l} x_1 = \arcsin{\frac{\sqrt{5} - 1}{2}} + 2\pi n\\ x_2 = \pi - \arcsin{\frac{\sqrt{5} - 1}{2}} + 2\pi k \end{array} \right.,\;n,k \in \mathbb{Z}.$

Mark these points on the unit circle.

Now let us turn to inequality $$\cos x \le 0.$$ The range of solutions to this inequality corresponds to the left semicircle. Therefore, the solution to the original equation is described by the formula

$x = \pi - \arcsin{\frac{\sqrt{5} - 1}{2}} + 2\pi k,\;k \in \mathbb{Z}.$

### Example 5.

Solve the equation

$\sin2x + \sqrt{2 - 5\sin^2x} = 0.$

Solution.

We write the equation in standard form and move on to the equivalent system:

$\sqrt{2 - 5\sin^2x} = -\sin2x \Leftrightarrow \left\{ \begin{array}{l} 2 - 5\sin^2x = \left({-\sin2x}\right)^2\\ -\sin2x \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 2 - 5\sin^2x = \sin^22x\\ \sin2x \le 0 \end{array} \right..$

Solve the resulting equation. Using the double-angle formula for sine and Pythagorean identity, we have

$\sin^22x = \left({2\sin x \cos x}\right)^2 = 4\sin^2x\cos^2x = 4\sin^2x\left({1 - \sin^2x}\right) = 4\sin^2x - 4\sin^4x.$

Plug in this expression into the equation:

$\sin^22x + 5\sin^2x - 2 = 0, \Rightarrow 4\sin^2x - 4\sin^4x + 5\sin^2x - 2 = 0, \Rightarrow 4\sin^4x - 9\sin^2x + 2 = 0.$

To solve this biquadratic equation, we make the substitution $$\sin^2x = t.$$ Then

$4t^2 - 9t + 2 = 0, \;D = \left({-9}\right)^2 - 4\cdot 4\cdot 2 = 49, \Rightarrow t_{1,2} = \frac{9 \pm \sqrt{49}}{4\cdot 2} = \frac{9 \pm 7}{8} = 2,\frac{1}{4}.$

There are no solutions for $$t_1 = 2:$$

$\sin^2x = t_1 = 2, \Rightarrow x = \varnothing.$

Find the values of $$x$$ at $$t_2 = \frac{1}{4}:$$

$\sin^2x = t_2 = \frac{1}{4}, \Rightarrow \sin x = \pm \frac{1}{2}, \Rightarrow x_1 = \pm\frac{\pi}{6} + 2\pi n, x_2 = \pm\frac{5\pi}{6} + 2\pi k,\;n,k \in \mathbb{Z}.$

Consider now the inequality $$\sin 2x \le 0:$$

$\Rightarrow -\pi + 2\pi m \le 2x \le 2\pi m,$
$\Rightarrow -\frac{\pi}{2} + \pi m \le x \le \pi m,\;m \in \mathbb{Z}.$

Draw all solutions on the unit circle.

It can be seen that points $${x = \frac{\pi}{6}, -\frac{5\pi}{6}}$$ do not fall into the allowed range of angles defined by inequality $$\sin2x \le 0.$$ So the solution of the equation looks like

$x = -\frac{\pi}{6} + \pi n,\;n \in \mathbb{Z}.$

### Example 6.

Solve the trigonometric equation

$4\sin x +\sqrt{6 - 6\tan^2x} = 0.$

Solution.

Bring the equation to standard form:

$\sqrt{6 - 6\tan^2x} = -4\sin x.$

This equation is equivalent to the following system:

$\sqrt{6 - 6\tan^2x} = -4\sin x \Leftrightarrow \left\{ \begin{array}{l} 6 - 6\tan^2x = \left({-4\sin x}\right)^2\\ -4\sin x \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 6 - 6\tan^2x - 16\sin^2x = 0\\ \sin x \le 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 8\sin^2x + 3\tan^2x - 3 = 0\\ \sin x \le 0 \end{array} \right..$

Rewrite the last equation in the following form:

$8\sin^2x + 3\frac{\sin^2x}{\cos^2x} - 3 = 0,$
$8\sin^2x\cos^2x + 3\sin^2x - 3\cos^2x = 0.$

Apply the double angle formulas:

$2\left({2\sin x\cos x}\right)^2 - 3\left({\cos^2x - \sin^2x}\right) = 0,$
$2\sin^22x - 3\cos2x = 0.$

Substitute $$\sin^22x = 1 - \cos^22x:$$

$2\left({1 - \cos^22x}\right) - 3\cos2x = 0,$
$2\cos^22x +3\cos2x - 2 = 0,$

Change the variable $$\cos2x = z$$ and solve the quadratic equation:

$2z^2 + 3z - 2 = 0,\;D = 3^2 - 4\cdot2\cdot\left({-2}\right) = 25, \Rightarrow z_{1,2} = \frac{-3 \pm \sqrt{25}}{2\cdot2} = \frac{-3 \pm 5}{4} = -2,\frac{1}{2}.$

For $$z_1 = -2$$ we have:

$\cos2x = z_1 = -2, \Rightarrow x = \varnothing.$

For $$z_2 = \frac{1}{2}:$$

$\cos2x = z_2 = \frac{1}{2}, \Rightarrow 2x = \pm\arccos\frac{1}{2} + 2\pi n, \Rightarrow 2x = \pm\frac{\pi}{3} + 2\pi n, \Rightarrow x = \pm\frac{\pi}{6} + \pi n,\;n \in \mathbb{Z}.$

The solution of the inequality $$\sin x \le 0$$ is obvious − it is represented by the lower semicircle.

So the solution of the equation on the unit circle are two points: $$x = -\frac{\pi}{6}$$ and $$x = -\frac{5\pi}{6}.$$ Both families of solutions can be represented by one formula:
$x = \left({-1}\right)^n \left({-\frac{\pi}{6}}\right) + \pi n = \left({-1}\right)^{n+1} \frac{\pi}{6} + \pi n,\;n \in \mathbb{Z}.$