## Task B-3.5.

For which real numbers $x$ does the function $f(x)=\sin x-\cos ^{2} x-1$ have the minimum value?
## Solution.

Let's rewrite the function $f(x)=\sin x-\cos ^{2} x-1$ in a convenient form.

$$
f(x)=\sin x-\cos ^{2} x-1=\sin x-\left(1-\sin ^{2} x\right)-1=\sin ^{2} x+\sin x-2
$$

With $t=\sin x$, the function $f(t)=t^{2}+t-2$ is a quadratic function that achieves its minimum value at $t_{0}=-\frac{1}{2}$.

Since $t=\sin x$, the given function will have its minimum value when $\sin x=-\frac{1}{2}$, (1 point) which corresponds to $x_{1}=\frac{7 \pi}{6}+2 k \pi, k \in \mathbb{Z}$ and $x_{2}=\frac{11 \pi}{6}+2 k \pi, k \in \mathbb{Z}$.