If $$2 \sin^2 q - 3 \sin q + 1 = 0, q$$ being positive angle less than or equal to 90 degrees, then the values of $$q$$ are

Given,

$$2 \sin^2 q - 3 \sin q + 1 = 0$$

$$=$$>  $$2\sin^2q-2\sin q-\sin q+1=0$$

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

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

$$=$$>  $$\sin q-1=0$$  or   $$2\sin q-1=0$$

$$=$$>  $$\sin q=1$$  or  $$\sin q=\frac{1}{2}$$

$$=$$>  $$\sin q=\sin90^{\circ\ }$$   or   $$\sin q=\sin30^{\circ\ }$$

$$=$$>  $$q=90^{\circ\ }$$   or   $$q=30^{\circ\ }$$

Hence, the correct answer is Option D