## Task 17/75

Given is the function

$$
y=2 \cos ^{2} x-3 \sqrt{3} \cos x-\sin ^{2} x+5
$$

We are looking for the values of $x$ for which $y$ assumes an extremum. The task is to solve this without using differential calculus!
Due to $\sin ^{2} x=1-\cos ^{2} x$, the given function can also be expressed in the form

$$
y=3 \cos ^{2} x-3 \sqrt{3} \cos x+4
$$

From this, we get

$$
y=\left(\sqrt{3} \cos x-\frac{3}{2}\right)^{2}-\frac{9}{4}+4=\left(\sqrt{3} \cos x-\frac{3}{2}\right)^{2}+\frac{7}{4}
$$

From this form of representation, one can immediately see:

1. The function value $y$ is minimal when the bracket expression is zero, i.e., for $x=2 k \pi \pm \frac{\pi}{6}$, it is $y_{\text {min }}=\frac{7}{4}$.
2. The function value $y$ is maximal when the absolute value of the bracket expression is maximal, i.e., for $x=\pi(2 k+1)$ and it is

$$
y_{\max }=\left(-\sqrt{3}-\frac{3}{2}\right)^{2}+\frac{7}{4}=7+3 \sqrt{3}
$$

In both cases, $k$ is any integer.