Question

Решительно плиз!!! cos(0,5пи+2x)+sinx=0 Sin^4x+cos2x=1

Answer 1

$\cos (0.5\pi +2x)+\sin x=0 \\ -\sin 2x+\sin x=0\\ -2\sin x\cos x+\sin x=0 \\ -\sin x(2\cos x-1)=0 \\ \left[\begin{array}{ccc}\sin x=0 \\ \cos x= \frac{1}{2} \end{array}\right\to \left[\begin{array}{ccc}x_1=\pi k,k \in Z\\ x_2=\pm \frac{\pi}{3}+2\pi n, n \in Z \end{array}\right$

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