Question

Please Help!
Find all solutions in the interval [0, 2π). 4 sin2 x - 4 sin x + 1 = 0

Answer 1

@zaynahf

Answer 2

Do you perhaps mean:\[4\sin^{2}x-4sinx+1=0\]?
In that case, you can treat it as an ordinary quadratic equation, with sinx as variable.
If you like, you can temporarily set p=sinx. The equation then becomes:\[4p^2-4p+1=0\]Then solve for p.
Later you can replace p with sinx again and solve for x.