Trigonometrical Equations 1 Question 14
14. The general value of $\theta$ satisfying the equation $2 \sin ^{2} \theta-3 \sin \theta-2=0$, is
(1995,2M)
(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) $n \pi+(-1)^{n} \frac{7 \pi}{6}$
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Answer:
Correct Answer: 14. (d)
Solution:
- Given, $2 \sin ^{2} \theta-3 \sin \theta-2=0$
$\Rightarrow \quad(2 \sin \theta+1)(\sin \theta-2)=0$
$ \begin{array}{ll} \Rightarrow \quad & \sin \theta=-1 / 2 \\ & {[\text { neglecting } \sin \theta=2, \text { as }|\sin \theta| \leq 1]} \end{array} $
$ \therefore \quad \theta=n \pi+(-1)^{n}(7 \pi / 6) $
