2cos^2(x-3pi/2)-sin(x-pi)=0 решите плиз!

[latex]2cos^2(x- \frac{3 \pi }{2} )-sin(x- \pi )=0 \\ 2sin^2x+sinx=0 \\ sinx(2sinx+1)=0 \\ \\ sinx=0 \\ x= \pi k \\ \\ 2sinx+1=0 \\ 2sinx=-1 \\ sinx=- \frac{1}{2} \\ x= \frac{3 \pi }{2} \pm \frac{\pi }{3} +2 \pi k= \left \{ {{ \frac{7 \pi }{6}+2 \pi k } \atop { \frac{11 \pi }{ 6} +2 \pi k}} \right. [/latex]