Problem 2. The number $x$ satisfies the condition $\frac{\sin 3 x}{\sin x}=\frac{5}{3}$. Find the value of the expression $\frac{\cos 5 x}{\cos x}$ for such $x$
Answer: $-\frac{11}{9}$.

Solution.

$$
\begin{aligned}
& \sin x \neq 0 \rightarrow \frac{\sin 3 x}{\sin x}=\frac{3 \sin x-4 \sin ^{3} x}{\sin x}=3-4 \sin ^{2} x \rightarrow 3-4 \sin ^{2} x=\frac{5}{3} \rightarrow \sin ^{2} x=\frac{1}{3} \rightarrow \\
& \rightarrow \cos ^{2} x=\frac{2}{3} \rightarrow \cos 2 x=\frac{1}{3} ; \\
& \frac{\cos 5 x}{\cos x}=\frac{\cos (3 x+2 x)}{\cos x}=\frac{\cos (3 x+2 x)}{\cos x}=\frac{\cos 3 x \cos 2 x-\sin 3 x \sin 2 x}{\cos x}= \\
& \frac{\left(4 \cos ^{3} x-3 \cos x\right) \cos 2 x-2 \sin 3 x \sin x \cos x}{\cos x}=\left(4 \cos ^{2} x-3\right) \cos 2 x-2 \sin 3 x \sin x= \\
& =(2 \cos 2 x-1) \cos 2 x-\cos 2 x+\cos 4 x==(2 \cos 2 x-1) \cos 2 x-\cos 2 x+2 \cos ^{2} 2 x-1= \\
& =\left(\frac{2}{3}-1\right) \cdot \frac{1}{3}-\frac{1}{3}+\frac{2}{9}-1=-\frac{11}{9} .
\end{aligned}
$$