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solve 2COS^2x+3sinx=0
2cos2x + 3sinx = 0
2(1-sin2x) +3sinx=0
2-2sin2x+3sinx=0
2sin2x-3sinx-2=0
now it is a quadratic equation
suppose sinx=y
2y 2 -3y-2=0
2y 2 -4y+y-2=0
2y(y-2)+1(y-2)=0
(y-2)(2y+1)=0

y=2 and -1/2
which implies
sinx=2 and sinx=-1/2
since sin x cannot be = 2
sinx=-1/2
sinx=sin210
x=210


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