Task B-3.2. Determine all solutions of the equation

$$
\frac{1}{2} \sin 2 x-\cos ^{2}\left(\frac{\pi}{2}-x\right)=0
$$
Solution. The given equation can be written in the form

$$
\begin{aligned}
\sin x \cos x-\sin ^{2} x & =0 \\
\sin x(\cos x-\sin x) & =0
\end{aligned}
$$

Then we have

$$
\sin x=0 \quad \text { or } \quad \cos x-\sin x=0
$$

From $\sin x=0$ it follows that $x=k \pi, k \in \mathbb{Z}$,

and from $\cos x-\sin x=0$, i.e., $\operatorname{tg} x=1$ it follows that $x=\frac{\pi}{4}+k \pi, k \in \mathbb{Z}$.