Question

2 cos квадрат x + sin 4x=1

Answer 1

$2\cos^2x+\sin4x=1\\ 2\cdot \frac{1+\cos2x}{2}+\sin 4x=1 \\ 1+\cos 2x+2\sin 2x\cos 2x=1 \\ \cos 2x+2\sin 2x\cos 2x=0 \\ \cos 2x(1+2\sin 2x)=0 \\ \left[\begin{array}{ccc}\cos 2x=0\\ \sin 2x=- \frac{1}{2} \end{array}\right \to \left[\begin{array}{ccc}2x= \frac{\pi}{2}+ \pi n,n \in Z \\ 2x=(-1)^{k+1} \cdot \frac{\pi}{6}+\pi k,k \in Z \end{array}\right\to \\ \to \left[\begin{array}{ccc}x= \frac{\pi}{4} + \frac{\pi n}{2} ,n \in Z \\ x=(-1)^{k+1}\cdot \frac{\pi}{12} + \frac{\pi k}{2}, k \in Z\end{array}\right$