11.1. Solve the equation $1-\cos 2 x+2 \cos ^{2} 2 x=4 \sin x \cdot \cos 2 x$.
Answer: $x=-\frac{\pi}{2}+2 \pi k, x=(-1)^{n} \frac{\pi}{6}+\pi n, n, k \in \mathbf{Z}$. Solution. The original equation is equivalent to $2\left(\sin ^{2} x+\cos ^{2} 2 x-2 \sin x \cdot \cos 2 x\right)=0 \Leftrightarrow(\sin x-\cos 2 x)^{2}=0 \Leftrightarrow \sin x-1+2 \sin ^{2} x=0 \Leftrightarrow$ $(\sin x)_{1,2}=\frac{-1 \pm 3}{4} \Leftrightarrow\left[\begin{array}{l}\sin x=-1 \\ \sin x=\frac{1}{2}\end{array}\right.$. From this, the result follows.