$5.50 \sin 6 x+2=2 \cos 4 x$.

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$5.50 \sin 6 x+2=2 \cos 4 x$.
5.50 We have

$\sin 6 x=2 \cos 4 x-2 ; \sin 6 x=-4 \sin ^{2} 2 x$.

Subtract $\sin 2 x$ from both sides of the equation:
$\sin 6 x-\sin 2 x=-4 \sin ^{2} 2 x-\sin 2 x$

$2 \sin 2 x \cos 4 x=-\sin 2 x(4 \sin 2 x+1)$.

Thus:

1) $\sin 2 x=0 ; 2 x=\pi k ; x_{1}=\frac{\pi k}{2}$
2) $2 \cos 4 x+4 \sin 2 x+1=0 ; 2-4 \sin ^{2} 2 x+4 \sin 2 x+1=0$; $4 \sin ^{2} 2 x-4 \sin 2 x-3=0, \sin 2 x=-\frac{1}{2}, \sin 2 x \neq \frac{3}{2} ;$

$x_{2}=(-1)^{k+1} \frac{\pi}{12}+\frac{\pi k}{2}$

Answer: $x_{1}=\frac{\pi k}{2}, x_{2}=(-1)^{k+1} \frac{\pi}{12}+\frac{\pi k}{2}, k \in Z$.