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:

  1. 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) $



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