## Task B-3.1.

Determine all solutions of the equation $\left|\cos ^{2} x-2 \sin x\right|=2$.
## Solution.

The expression inside the absolute value can be equal to 2 or -2.

In the first case, if $\cos ^{2} x-2 \sin x=2$, then:

$$
\begin{aligned}
1-\sin ^{2} x-2 \sin x-2=0, & 1 \text { point } \\
(\sin x+1)^{2}=0, & \\
\sin x=-1, & 1 \text { point }
\end{aligned}
$$

In the second case, if $\cos ^{2} x-2 \sin x=-2$, then:

$$
\begin{gathered}
1-\sin ^{2} x-2 \sin x+2=0 \\
\sin ^{2} x+2 \sin x-3=0
\end{gathered}
$$

Solving the quadratic equation, we get

$$
\sin x=-3
$$

which is impossible and

$$
\sin x=1
$$

from which

$$
x_{2}=\frac{\pi}{2}+2 k \pi, k \in \mathbb{Z}
$$

Note: The solution can (but does not have to) be written as a single set:

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