## Task B-3.2.

Solve the inequality $|\cos x| \leqslant \cos x + 2 \sin x$ on the interval $[0, 2\pi\rangle$.
## Solution.

Let's solve the inequality on the given interval by breaking it down into two cases.

1. case: $\cos x \geqslant 0$

$\cos x \leqslant \cos x+2 \sin x$

$\sin x \geqslant 0$

Given the condition $\cos x \geqslant 0$, it follows that $x \in\left[0, \frac{\pi}{2}\right]$.

2. case: $\cos x<0$

$-\cos x \leqslant \cos x+2 \sin x$

$-2 \sin x \leqslant 2 \cos x /:(-2)$

$\sin x \geqslant-\cos x /: \cos x<0$

$\operatorname{tg} x \leqslant-1$

Given the condition $\cos x<0$, it follows that $x \in\left\langle\frac{\pi}{2}, \frac{3 \pi}{4}\right]$.

The solution set of the original inequality is the union of the solutions obtained in the two cases:

$$
x \in\left[0, \frac{3 \pi}{4}\right]
$$