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
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
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.
\[2 + \sin x = t_2 = 1, \Rightarrow \sin x = -1, \Rightarrow x = -\frac{\pi}{2} + 2\pi n,\;n\in\mathbb{Z}.\]
\[\cos x = 1, \Rightarrow x_2 = 2\pi k,\;k\in\mathbb{Z}.\]
Mark these points on the unit circle.
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:
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
\[\Rightarrow -\frac{\pi}{2} + \pi m \le x \le \pi m,\;m \in \mathbb{Z}.\]
Draw all solutions on the unit circle.
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
\[\cos2x = z_1 = -2, \Rightarrow x = \varnothing.\]
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.
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:
