Question

)1)2sin²x-2sinx=1 2)2sin²x+5cosx-4=0

Answer 1

$1)\; 2sin^2x-2sinx=1\\\\t=sinx,\; 2t^2-2t-1=0\\\\\frac{D}{4}=1+2=3,\; t_1=\frac{1-\sqrt3}{2},\; t_2=\frac{1+\sqrt3}{2}1\; (ne\; podxodit)\\\\sinx=\frac{1-\sqrt3}{2},\; x=(-1)^{n}arcsin\frac{1-\sqrt3}{2}+\pi n,\; n\in Z\\\\2)\; 2sin^2x+5cosx-4=0\\\\2(1-cos^2x)+5cosx-4=0\\\\2cos^2x-5cosx+2=0\\\\t=cosx,\; 2t^2-5t+2=0$

$D=25-16=9,t_1=\frac{5-3}{4}=\frac{1}{2},\; t_2=21\\\\cosx=\frac{1}{2},\; x=\pm \frac{\pi}{3}+2\pi n,\; n\in Z$