Problem 2. (2 points)

$\sin x \sin 3 x=\frac{5}{16}$. Find $\cos x \cos 3 x$.

If there are multiple possible answers, write them separated by a semicolon.
Answer: $7 / 16 ;-9 / 16$

## Solution:

General form of the problem: $\sin x \sin 3 x=\frac{5}{16}=\alpha$.

$\alpha=\sin x \sin 3 x=\sin x\left(3 \sin x-4 \sin ^{3} x\right)=3 \sin ^{2} x-4 \sin ^{4} x$.

Let $\sin ^{2} x$ be $t$, and we get a quadratic equation in $t: 4 t^{2}-3 t+\alpha=0$, from which $t=$ $\frac{3 \pm \sqrt{9-4 \alpha}}{8}$.

On the other hand, $\cos x \cos 3 x=\cos x\left(4 \cos ^{3} x-3 \cos x=4 \cos ^{4} x-3 \cos ^{2} x\right)$. Considering that $\cos ^{2} x=$ $1-\sin ^{2} x=1-t$, we get that $\cos x \cos 3 x=4(1-t)^{2}-3 t=4 t^{2}-11 t+4=\alpha-8 t+4$. Substituting the value of $\alpha$, obtained by solving the quadratic equation, we get the final answer.