Solve the following equations: 0<x<2pi?

Question

cos2x=7sinx-3

2sin^2x=3cosx

Solve the folowing 0 2sin3x= -sq root 3

Answer 1

cos^2(x) -sin^2(x) =7 sin x -3
1-2 sin^2(x) =7sin x-3
call sin x= z
2z^2+7z-4=0
z=(( -7+-sqrt(49+32)/4 = (-7+-9)/4
z=-4 impossible)
z=1/2 so x= pi/6 and 5pi/6
sin3x= -1/2sqrt3 3x= -pi/3+2k*pi and x = -pi/9 +2kpi/3
for k=1 x= 5pi/9(ok) solution
for k= 2 x= 11pi/9 (outside)