## Task 12/70

Without using differential calculus, determine the smallest and largest value of the function

$$
y=f(x)=2-2 \cos ^{2} x-\sin x
$$
The term $2-2 \cos ^{2} x-\sin x$ is transformed using known trigonometric relationships:

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

The maximum value is clearly achieved for $\sin x=-1$, since then $\sin ^{2} x=1$ and $-\sin x=1$. Thus, $f_{\max }(x)=+3$.

To determine the minimum function value, further transformations are performed:

$$
y=f(x)=2 \sin ^{2} x-\sin x=2\left(\sin ^{2} x-\frac{1}{2} \sin x+\frac{1}{16}\right)-\frac{1}{8}=2\left(\sin x-\frac{1}{4}\right)^{2}-\frac{1}{8}
$$

The minimum value occurs when $\sin x-\frac{1}{4}=0$. Therefore, $f_{\min }=-\frac{1}{8}$.