Trigonometrical Equations 1 Question 3
3. Let S= {${\theta \in[-2 \pi, 2 \pi]: 2 \cos ^2 \theta+3 \sin \theta=0}$}, then the sum of the elements of $S$ is
(2019 Main, 9 April I)
(a) $2 \pi$
(b) $\pi$
(c) $\frac{5 \pi}{3}$
(d) $\frac{13 \pi}{6}$
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Answer:
Correct Answer: 3. (a)
Solution:
- We have, $\theta \in[-2 \pi, 2 \pi]$
$$ \begin{aligned} & \text { and } \quad 2 \cos ^{2} \theta+3 \sin \theta=0 \\ & \Rightarrow \quad 2\left(1-\sin ^{2} \theta\right)+3 \sin \theta=0 \\ & \Rightarrow \quad 2-2 \sin ^{2} \theta+3 \sin \theta=0 \\ & \Rightarrow \quad 2 \sin ^{2} \theta-3 \sin \theta-2=0 \\ & \Rightarrow \quad 2 \sin ^{2} \theta-4 \sin \theta+\sin \theta-2=0 \\ & \Rightarrow 2 \sin \theta(\sin \theta-2)+1(\sin \theta-2)=0 \\ & \Rightarrow \quad(\sin \theta-2)(2 \sin \theta+1)=0 \\ & \therefore \quad \sin \theta=\frac{-1}{2} \\ & {[\because(\sin \theta-2) \neq 0]} \\ & \therefore \quad \theta=2 \pi-\frac{\pi}{6},-\pi+\frac{\pi}{6},-\frac{\pi}{6}, \pi+\frac{\pi}{6} \\ & {[\because \theta \in[-2 \pi, 2 \pi]]} \end{aligned} $$
Now, sum of all solutions
$$ =2 \pi-\frac{\pi}{6}-\pi+\frac{\pi}{6}-\frac{\pi}{6}+\pi+\frac{\pi}{6}=2 \pi $$
