4. Determine how many roots the equation has on the interval $[-\pi ; \pi]$

$$
\frac{2 \cos 4 x+1}{2 \cos x-\sqrt{3}}=\frac{2 \sin 4 x-\sqrt{3}}{2 \sin x-1}
$$

and specify these roots
# Problem 4.

Answer: 4 roots: $-5 \pi / 6, -\pi / 3, \pi / 2, 2 \pi / 3$.

Solution. Admissible values are determined by the conditions: $\cos x \neq \frac{\sqrt{3}}{2}, \sin x \neq \frac{1}{2}$, and therefore on the interval $[-\pi, \pi]: x \neq \pm \frac{\pi}{6}, x \neq \frac{5 \pi}{6}$.

Rewrite the equation as

$$
\begin{gathered}
(2 \cos 4 x+1)(2 \sin x-1)=(2 \sin 4 x-\sqrt{3})(2 \cos x-\sqrt{3}) \Longleftrightarrow \\
\Longleftrightarrow 4 \sin 3 x-4 \sin \left(x+\frac{\pi}{3}\right)-4 \sin \left(4 x-\frac{\pi}{6}\right)+4=0
\end{gathered}
$$

The substitution $y=x+\frac{\pi}{3}$ leads to the equation

$$
\sin 3 y+\sin y+\cos 4 y-1=0 \Longleftrightarrow \sin ^{2} y \cos y(1-2 \sin y)=0
$$

from which $\sin y=0$ or $\cos y=0$ or $\sin y=1 / 2$. We obtain three series of solutions: $x=-\frac{\pi}{3}+\frac{\pi n}{2}$, $x=-\frac{\pi}{6}+2 \pi n, x=\frac{\pi}{2}+2 \pi n, n \in \mathbb{Z}$. Direct selection gives four roots on the interval $[-\pi, \pi]:-\frac{5 \pi}{6},-\frac{\pi}{3}, \frac{\pi}{2}, \frac{2 \pi}{3}$.