We will factorise the equation : 2(sin^2)x + 5sinx-3 =0
2(sin^2)x + 6sinx-sinx -3= 2sinx(sinx+3)-1(sinx+3)=0
Therefore we get 2 values of sinx,
sinx=1/2, -3
Since sinx can only lie in the interval [-1,1], we will only consider 1/2 as the solution, therefore number of solutions=1
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