5.10 $1-\cos 6 x=\tan 3 x$.

5.11 $1+\sin 2 x=(\cos 3 x+\sin 3 x)^{2}$.
5.10 We have

$2 \sin ^{2} 3 x=\operatorname{tg} 3 x, \cos 3 x \neq 0 ; 2 \sin ^{2} 3 x=\frac{\sin 3 x}{\cos 3 x} ；$

$2 \sin ^{2} 3 x \cos 3 x=\sin 3 x$

1) $\sin 3 x=0,3 x=\pi k ; x_{1}=\frac{\pi k}{3}$;
2) $2 \sin 5 x \cos 3 x=1 ; \sin 6 x=1,6 x=\frac{\pi}{2}+2 \pi k ; x_{2}=\frac{\pi}{12}(4 k+1)$.

Solution: $\quad x_{1}=\frac{\pi k}{3}, x_{2}=\frac{\pi}{12}(4 k+1), k \in Z$.