8.147. $\cos x - \sin x = 4 \cos x \sin^2 x$.
Solution.

By the formulas for reducing the degree

$\cos x-\sin x=2 \cos x(1-\cos 2 x) \Leftrightarrow 2 \cos x \cos 2 x-(\cos x+\sin x)=0$, $2 \cos x(\cos x+\sin x)(\cos x-\sin x)-(\cos x+\sin x)=0$, $(\cos x+\sin x)(2 \cos x(\cos x-\sin x)-1)=0, \quad(\cos x+\sin x) \times$ $\times\left(2 \cos ^{2} x-2 \sin x \cos x-1\right)=0 \Leftrightarrow(\cos x+\sin x)(\cos 2 x-\sin 2 x)=0$.

Then:

1) $\cos x+\sin x=0 \Leftrightarrow \operatorname{tg} x=-1, \quad x_{1}=-\frac{\pi}{4}+\pi k=\frac{\pi}{4}(4 k-1), k \in Z$;
2) $\cos 2 x-\sin 2 x=0 \Leftrightarrow \operatorname{tg} 2 x=1, x_{2}=\frac{\pi}{8}+\frac{\pi k}{2}=\frac{\pi}{8}(4 n+1), n \in Z$.

Answer: $\quad x_{1}=\frac{\pi}{4}(4 k-1), x_{2}=\frac{\pi}{8}(4 n+1), k, n \in Z$.