Question

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

• A. $n\pi+(-1)^n\frac{\pi}{6}$
• B. $n\pi+(-1)^n\frac{\pi}{2}$
• C. $n\pi+(-1)^n\frac{5\pi}{6}$
• D. $n\pi+(-1)^n\frac{7\pi}{6}$

Answer 1

Answer: (D)

$n\pi+(-1)^n\frac{7\pi}{6}$

Answer 2

$2 \sin^2 \, \theta - 3 \sin \theta - 2 = 0$ $\Rightarrow \, \, \, (\sin \theta - 2) (2 \sin \theta + 1) = 0$ $\Rightarrow \, \, \, \sin \theta = 2$ or $\sin \theta = - \frac{1}{2}$ But $\sin \theta = 2$ is not possible. $\therefore \, \, \, \, \, \sin \theta = - \frac{1}{2} = \sin \left( \frac{7 \pi }{ 6} \right)$ $\therefore \, \, \, \, \theta = n \pi + (-1)^n \left( \frac{7 \pi }{6} \right)$