[Trigonometric Identities.]

Find $\frac{\sin 5 x}{\sin x}$, if $\frac{\sin 3 x}{\sin x}=\frac{6}{5}$.

#
$11 / 5 \sin x=\sin 3 x+\sin x=2 \sin 2 x \cos x=4 \sin x \cos ^{2} x$, so, $\cos ^{2} x=11 / 20$.

Therefore, $\sin 5 x+\sin x=2 \sin 3 x \cos 2 x=12 / 5 \sin x\left(2 \cos ^{2} x-1\right)=2.4 \cdot(1.1-1) \sin x=0.24 \sin x$, and $\sin 5 x=$ $-0.76 \sin x$

## Answer

$-0.76$.