Question

Как решить вот это уравнение? 2cos^2x +sin4x=1?

Answer 1

$2cos^{2}x+sin4x=1 \\ sin4x=cos^{2}x +sin^{2}x - 2cos^{2}x \\ sin4x=sin^{2}x-cos^{2}x \\ sin4x=-(cos^{2}x-sin^{2}x) \\ sin4x=-cos2x \\ sin4x +cos2x=0 \\ 2 \cdot sin2x \cdot cos2x+cos2x=0 \\ cos2x \cdot (2 \cdot sin2x+1)=0 \\ cos2x = 0...................sin2x=- \frac{1}{2} \\ 2x_{n}= \frac{\pi}{2}+\pi n, n \epsilon Z.....2x_{k}=- \frac{\pi}{6} +2\pi k, k \epsilon Z \wedge2x_{k}=- \frac{5\pi}{6} +2\pi k, k \epsilon Z$
$x_{n}= \frac{\pi}{4}+\frac{\pi n}{2}, n \epsilon Z.....x_{k}=- \frac{\pi}{12} +\pi k, k \epsilon Z \wedge x_{k}=- \frac{5\pi}{12} +\pi k, k \epsilon Z$