Помогите решить срочно очень!!!!!cos2x-3sinx=-1

$cos2x-3sinx+1=0\\cos ^{2} x-sin ^{2} x-3sinx+1=0 \\ 1-sin ^{2} x-sin ^{2} x-3sinx+1=0\\-2sin ^{2} x-3sinx+2=0/*(-1)\\2sin ^{2} x+3sinx-2=0\\sinx=t\\-1 \leq t \leq 1 \\ 2t ^{2} +3t-2=0\\D=9+4*4=9+16=25\\ \sqrt{D} =5\\t _{1} = \frac{-3+5}{4} = \frac{2}{4} = \frac{1}{2} \\t _{2} = \frac{-3-5}{4} =-2\\\\sinx= \frac{1}{2} \\\\x=(-1) ^{n} \frac{ \pi }{6} + \pi n$
n ∈ Z