Question

помогите решить
sin(P-2x) =cos^2x [-P/2;3P/2)]​

Answer 1

Ответ:

$\sin(\pi - 2x) = { \cos }^{2} (x) \\ \sin(2x) - { \cos}^{2} (x) = 0 \\ \\ 2 \sin(x) \cos(x) - { \cos }^{2}(x) = 0 \\ \cos(x) (2 \sin(x) - 1) = 0 \\ \\ \cos(x) = 0 \\ x1 = \frac{\pi}{2} + \pi \: n \\ \\ \sin(x) = \frac{1}{2} \\ x2 = \frac{\pi}{6} + 2\pi \: n \\ x3 = \frac{5\pi}{6} + 2 \pi \: n$

n принадлежит Z.

Ответ:

$- \frac{\pi}{2} ; \frac{\pi}{6} ; \frac{\pi}{2} ;\frac{5\pi}{6} ; \frac{3\pi}{2}$