Question

1)tg^2x=3 2)2sin^2x-2sinx=1 3)2sin^2x+5cosx - 4=0

Answer 1

$1)\; tg^2x=3\\\\tgx=\pm \sqrt3\\\\tgx=\sqrt3,\; x=\frac{\pi}{3}+\pi n,n\in Z\\\\tgx=-\sqrt3,\; x=-\frac{\pi }{3}+\pi n,n\in Z\\\\2)\; 2sin^2x-2sinx=1\\\\t=sinx,\; 2t^2-2t-1=0\\\\D=4+8=12,\sqrt{D}=2\sqrt3,\\\\t_1=\frac{2-2\sqrt3}{4}=\frac{1-\sqrt3}{2},\; t_2=\frac{1+\sqrt3}{2}\approx 1,351$

$sinx=\frac{1-\sqrt3}{2},\; x=(-1)^{n}arcsin\frac{1-\sqrt3}{2}+\pi n,n\in Z$

$3)\; 2sin^2x+5cosx-4=0\\\\2(1-cos^2x)+5cosx-4=0\\\\2cos^2x-5cosx+2=0\\\\t=cosx,\; 2t^2-5t+2=0\\\\D=25-16=9\\\\t_1=\frac{5-3}{4}=\frac{1}{2},\; t_2=21\\\\cosx=\frac{1}{2},\; x=\pm \frac{\pi}{3}+2\pi n,\; n\in Z$