Precalculus

Trigonometry

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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.

Solutions of the equation sinx = sqrt(1-cosx)
Figure 1.

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.

Solutions of the equation sqrt(sinx) + cosx = 0
Figure 2.

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.

Solutions of the equation sin2x + sqrt(2-5sin^2x) = 0
Figure 3.

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.

Solutions of the equation 4sinx + sqrt(6-6tan^2x) = 0
Figure 4.

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:

\[x = \left({-1}\right)^n \left({-\frac{\pi}{6}}\right) + \pi n = \left({-1}\right)^{n+1} \frac{\pi}{6} + \pi n,\;n \in \mathbb{Z}.\]