If $2 \cos ^2 x+3 \sin x-3=0,0 \leq x \leq 180^{\circ}$, then the value of x is
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If $2 \cos ^2 x+3 \sin x-3=0,0 \leq x \leq 180^{\circ}$, then the value of $x$ is

(A) $30^{\circ}, 90^{\circ}, 150^{\circ}$

(B) $60^{\circ}, 120^{\circ}, 180^{\circ}$

(C) $0^{\circ}, 30^{\circ}, 150^{\circ}$

(D) $45^{\circ}, 90^{\circ}, 135^{\circ}$

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Best answer

SOLUTION —

$\text { }  \because  2 \cos ^2 x+3 \sin x-3  =0$

$\Rightarrow  2-2 \sin ^2 x+3 \sin x-3  =0$

$\Rightarrow  (2 \sin x-1)(\sin x-1) =0$

$\Rightarrow  \sin x=\frac{1}{2} \text { or } \sin x=1$

$\Rightarrow  x=\frac{\pi}{6}, \frac{5 \pi}{6} \text { and } \frac{\pi}{2} \Rightarrow 30^{\circ}, 150^{\circ} \text { and } 90^{\circ}$

So, The correct option will be (A).

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