The value of $$x$$, if $$2 \sin^2 x = 2 - 3 \sin x$$, is:

Given,  $$2 \sin^2 x = 2 - 3 \sin x$$

$$=$$>  $$2\sin^2x+3\sin x-2=0$$

$$=$$>  $$2\sin^2x-\sin x+4\sin x-2=0$$

$$=$$>  $$\sin x\left(2\sin x-1\right)+2\left(2\sin x-1\right)=0$$

$$=$$>  $$\left(2\sin x-1\right)\left(\sin x+2\right)=0$$

$$=$$>  $$2\sin x-1=0$$  or  $$\sin x+2=0$$

$$=$$>  $$\sin x=\frac{1}{2}$$  or  $$\sin x=-2$$

$$\sin x=-2$$ is not possible

$$=$$>  $$\sin x=\frac{1}{2}$$

$$=$$>  $$\sin x=\sin\frac{\pi\ }{6}$$

$$=$$>  $$x=\frac{\pi\ }{6}$$

Hence, the correct answer is Option A