Question : The value of $\theta$ $ \left ( 0\leq \theta \leq 90^{\circ} \right )$ satisfying $2\sin^{2}\theta = 3\cos \theta$ is:
Option 1: $60^{\circ}$
Option 2: $30^{\circ}$
Option 3: $90^{\circ}$
Option 4: $45^{\circ}$
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Correct Answer: $60^{\circ}$
Solution :
Given that, $2\sin^{2}\theta = 3\cos\theta$.
Use the identity $\sin^{2}\theta + \cos^{2}\theta = 1$
$⇒2(1-\cos^{2}\theta) = 3\cos\theta$
$⇒2\cos^{2}\theta + 3\cos\theta - 2 = 0$
$⇒(2\cos\theta - 1)(\cos\theta + 2) = 0$
$⇒\cos\theta = \frac{1}{2}$ or $\cos\theta = -2$
Since $0^{\circ} \leq \theta \leq 90^{\circ}$, we can take $\cos\theta = \frac{1}{2}$.
$⇒\cos\theta = \frac{1}{2}=\cos60^{\circ}$
$\Rightarrow \theta = 60^{\circ}$
Hence, the correct answer is $60^{\circ}$.
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