Question

Решите уравнение:
Sin2x+2cox^2x+cos2x=0

Answer 1

$sin2x+2cos^2x+cos2x=0\ sin2x+1+cos2x+cos2x=0\ sin2x+2cos2x=-1\ sqrt{5} sin(2x+arcsin frac{2}{sqrt{5} } )=-1\ \ 2x+arcsin frac{2}{sqrt{5} } =(-1)^{k+1}cdot arcsin frac{1}{sqrt{5} } + pi k,k in mathbb{Z}\ \ boxed{x=(-1)^{k+1}cdot0.5arcsin frac{1}{sqrt{5} }-0.5arcsin frac{2}{sqrt{5} } +0.5 pi k,k in mathbb{Z} }$