Question

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

Answer 1

$1); tg^2x=3\\tgx=pm sqrt3\\tgx=sqrt3,; x=frac{pi}{3}+pi n,nin Z\\tgx=-sqrt3,; x=-frac{pi }{3}+pi n,nin Z\\2); 2sin^2x-2sinx=1\\t=sinx,; 2t^2-2t-1=0\\D=4+8=12,sqrt{D}=2sqrt3,\\t_1=frac{2-2sqrt3}{4}=frac{1-sqrt3}{2},; t_2=frac{1+sqrt3}{2}approx 1,35>1$

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

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