Question

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

Answer 1

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