Let S=θ∈[−2π,2π]:2cos2⁡θ+3sin⁡θ=0Then the sum of the elements of S is :

# Let $S=\left\{\theta \in \left[-2\pi ,2\pi \right]:2{\mathrm{cos}}^{2}\theta +3\mathrm{sin}\theta =0\right\}$Then the sum of the elements of is :

1. A

$2\pi$

2. B

$\pi$

3. C

$5\pi /3$

4. D

$13\pi /6$

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### Solution:

Given, $2{\mathrm{cos}}^{2}\theta +3\mathrm{sin}\theta =0$

$\begin{array}{l}⇒2\left(1-{\mathrm{sin}}^{2}\theta \right)+3\mathrm{sin}\theta =0\\ ⇒2-2{\mathrm{sin}}^{2}\theta +3\mathrm{sin}\theta =0\\ ⇒2{\mathrm{sin}}^{2}\theta -4\mathrm{sin}\theta +\mathrm{sin}\theta -2=0\\ ⇒2\mathrm{sin}\theta \left(\mathrm{sin}\theta -2\right)+1\left(\mathrm{sin}\theta -2\right)=0\\ ⇒\left(2\mathrm{sin}\theta +1\right)\left(\mathrm{sin}\theta -2\right)=0\end{array}$

$⇒\mathrm{sin}\theta =-\frac{1}{2}$

Required sum $=\left[\frac{-\pi -5\pi +7\pi +11\pi }{6}\right]=\frac{12\pi }{6}=2\pi$

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