$5.14 \sin 3 x+\sin 5 x=2\left(\cos ^{2} 2 x-\sin ^{2} 3 x\right)$.
5.14 To the left side of the equation, we apply formula (4.19), and to the right side, we apply formulas (4.17) and (4.16):

$2 \sin 4 x \cos x=2\left(\frac{1+\cos 4 x}{2}-\frac{1-\cos 6 x}{2}\right)$.

$2 \sin 4 x \cos x=\cos 4 x+\cos 6 x ; \sin 4 x \cos x=\cos 5 x \cos x$

1) $\cos x=0 ; x_{1}=\frac{\pi}{2}+\pi k$
2) $\cos 5 x=\sin 4 x ; \cos 5 x-\cos \left(\frac{\pi}{2}-4 x\right)=0$;

$\sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \sin \left(\frac{\pi}{4}-\frac{9 x}{2}\right)=0$; from this we get:

a) $\frac{\pi}{4}+\frac{x}{2}=\pi k, x=-\frac{\pi}{2}+2 \pi k-$ these values are included in the solution $x_{1}$;

b) $\frac{9 x}{2}-\frac{\pi}{4}=\pi k, 9 x=\frac{\pi}{2}+2 \pi k ; x_{2}=\frac{\pi}{18}+\frac{2 \pi k}{9}$.

Solution: $\quad x_{1}=\frac{\pi}{2}(2 k+1), x_{2}=\frac{\pi}{18}(4 k+1), k \in Z$.