4. Solve the equation $\operatorname{ctg} 2 x \sqrt{\sin x \cos x}-\sqrt{1-\sin x \cos x}=0$.
# Solution:

$\operatorname{ctg} 2 x \sqrt{\sin x \cos x}=\sqrt{1-\sin x \cos x}$. Given $\operatorname{ctg} 2 x \geq 0$, we square both sides of the equation ($1-\sin x \cos x>0$ for all $x$): $\operatorname{ctg}^{2} 2 x \sin x \cos x=1-\sin x \cos x \Leftrightarrow$ $\operatorname{ctg}^{2} 2 x \sin 2 x=2-\sin 2 x \Leftrightarrow \sin 2 x \neq 0, \quad \cos ^{2} 2 x=\sin 2 x(2-\sin 2 x) \Leftrightarrow$ $\sin 2 x(2-\sin 2 x)=1-\sin ^{2} 2 x \Leftrightarrow \sin 2 x=\frac{1}{2} \Leftrightarrow 2 x=(-1)^{n} \frac{\pi}{6}+\pi n, n \in Z$. Considering the condition $\operatorname{ctg} 2 x \geq 0$, we finally have $x=\frac{\pi}{12}+\pi n, n \in Z$.

Answer: $x=\frac{\pi}{12}+\pi n, n \in Z$.