Question

решить уравнение sinX+2cos2X=3/2

Answer 1

$\sin x+2\cos 2x=1.5~~|\cdot 2\\ 2\sin x+4\cos 2x=3\\ 2\sin x+4(1-2\sin^2x)=3\\ 2\sin x+4-8\sin^2x-3=0\\ 8\sin^2x-2\sin x-1=0\\ D=b^2-4ac=(-2)^2-4\cdot 8\cdot (-1)=36$

$\sin x=-0.25~~~\Rightarrow~~~ \boxed{x=(-1)^{k+1}\arcsin0.25+\pi k,k \in \mathbb{Z}} \\ \\ \sin x=0.5~~~\Rightarrow~~~ \boxed{x=(-1)^k\cdot\frac{\pi}{6}+\pi k,k \in \mathbb{Z}}$