8.233. $\sin 2 x-2 \cos ^{2} x+4(\sin x-\cos x+\operatorname{tg} x-1)=0$.

8.233. $\sin 2 x-2 \cos ^{2} x+4(\sin x-\cos x+\tan x-1)=0$.
## Solution.

Domain of definition: $\cos x \neq 0$.

Let's write the equation in the form

$$
\begin{aligned}
& \left(2 \sin x \cos x-2 \cos ^{2} x\right)+4\left((\sin x-\cos x)+\frac{\sin x}{\cos x}-1\right)=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos x(\sin x-\cos x)+4\left((\sin x-\cos x)+\frac{\sin x-\cos x}{\cos x}\right)=0 \Leftrightarrow
\end{aligned}
$$

$$
\begin{aligned}
& 2 \cos x(\sin x-\cos x)+4(\sin x-\cos x)\left(1+\frac{1}{\cos x}\right)=0 \Leftrightarrow \\
& \Leftrightarrow 2(\sin x-\cos x)\left(\cos x+2+\frac{2}{\cos x}\right)=0
\end{aligned}
$$

## From this

1) $\sin x-\cos x=0$
2) $\cos x+2+\frac{2}{\cos x}=0$. Dividing the first equation by $\cos x \neq 0$, we find $\operatorname{tg} x=1, x_{1}=\frac{\pi}{4}+\pi k=\frac{\pi}{4}(4 k+1), k \in Z$. Multiplying the second equation by $\cos x \neq 0$, we get $\cos ^{2} x+2 \cos x+2 \neq 0(D<0), \varnothing$.

Answer: $x=\frac{\pi}{4}(4 k+1), k \in Z$.