4 sin ^ 2 x- 4 sin x + 1 = 0
\(4\sin^2x-4\sin x+1=(2\sin x-1)^2=0\implies \sin x=\dfrac12\)This happens for \(x=\dfrac\pi6+2n\pi\) and \(x=\dfrac{5\pi}6+2n\pi\) where \(n\) is any integer.
