Task B-3.2. Determine the minimum and maximum value of the function

$$
f(x)=2 \sin x-\cos (2 x)
$$
## Solution.

Let's rewrite the function $f$ in a different form,

$$
\begin{aligned}
f(x) & =2 \sin x - \cos (2 x) \\
& =2 \sin x - \left(\cos ^{2} x - \sin ^{2} x\right) = \\
& =2 \sin x - \left(1 - 2 \sin ^{2} x\right) = \\
& =2 \sin ^{2} x + 2 \sin x - 1 = \\
& =2\left(\sin x + \frac{1}{2}\right)^{2} - \frac{3}{2} .
\end{aligned}
$$

The minimum value of the function $f$ is $-\frac{3}{2}$, and it is achieved when $\sin x = -\frac{1}{2}$. The maximum value of the function $f$ is for $\sin x = 1$ and is $2 \cdot \frac{9}{4} - \frac{3}{2} = 3$.

(3 points)