NIMCET 2011 — Mathematics PYQ

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Question

NIMCET 2011 — Mathematics PYQ

NIMCET | Mathematics | 2011

The general value of $\theta$ satisfying the equation $2 \sin^2 \theta - 3 \sin \theta - 2 = 0$ is:

Choose the correct answer:

Explanation

1. Factorize the quadratic equation:

Let $\sin \theta = t$ . The equation becomes $2t^2 - 3t - 2 = 0$ .

2t^2 - 4t + t - 2 = 0

2t(t - 2) + 1(t - 2) = 0

(2t + 1)(t - 2) = 0

2. Solve for $t$ :

$t = 2 \implies \sin \theta = 2$ (Not possible as $|\sin \theta| \le 1$ )
$2t + 1 = 0 \implies t = -\frac{1}{2} \implies \sin \theta = -\frac{1}{2}$

3. Find the general solution:

$\sin \theta = -\frac{1}{2} = \sin(-\frac{\pi}{6})$

The general solution for $\sin \theta = \sin \alpha$ is $\theta = n\pi + (-1)^n \alpha$ .

\theta = n\pi + (-1)^n (-\frac{\pi}{6})

\theta = n\pi + (-1)^{n+1} \frac{\pi}{6}

This is equivalent to $\theta = n\pi + (-1)^n \frac{7\pi}{6}$ as $\sin(\frac{7\pi}{6}) = -\frac{1}{2}$ .

Correct Option: (d)