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Let $S=\left[ \theta \in \left[ -2\pi ,2\pi \right]:2{{\cos }^{2}}\theta +3\sin \theta =0 \right]$ . Then the sum of the elements of $S$ is?
A. $\dfrac{13\pi }{6}$
B. $0$
C. $2\pi $
D. $\dfrac{5\pi }{3}$

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Last updated date: 13th Jun 2024
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Answer
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Hint: We will first convert the given equation in quadratic form, we will do this by writing \[{{\cos }^{2}}\theta \] as $(1-{{\sin }^{2}}\theta )$, we will then substitute $\sin \theta $ as $m$, in this way calculation will become easier. We will then find the solutions of the quadratic equation by factorizing the equation, after finding the solutions we will again re-substitute $\sin \theta $and then find the inverse in order to find out the elements of $S$.

Complete step-by-step answer:
We are given the range of $\theta $ as $\left[ -2\pi ,2\pi \right]$ and $2{{\cos }^{2}}\theta +3\sin \theta =0$, We will start by solving for $\theta $:
We have with us: $2{{\cos }^{2}}\theta +3\sin \theta =0\text{ }........\text{ Equation 1}\text{.}$
Now we know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $
Now putting this value in our equation no. 1
\[\begin{align}
  & 2{{\cos }^{2}}\theta +3\sin \theta =0 \\
 & \Rightarrow 2\left( 1-{{\sin }^{2}}\theta \right)+3\sin \theta =0 \\
 & \Rightarrow 2-2si{{n}^{2}}\theta +3sin\theta =0 \\
\end{align}\]
Taking the elements of L.H.S. to R.H.S. we will get the following equation: \[2si{{n}^{2}}\theta -3sin\theta -2=0\]
Now we will solve this quadratic equation by factorization method but let us first substitute $\sin \theta $ as $m$, in this way our calculation will become easy:
Let $m=\sin \theta $, therefore \[2si{{n}^{2}}\theta -3sin\theta -2=0\] will become : \[2{{m}^{2}}-3m-2=0\]. We will start by factorizing the equation:
 \[\begin{align}
  & \Rightarrow 2{{m}^{2}}-3m-2=0 \\
 & \Rightarrow 2{{m}^{2}}-4m+m-2=0 \\
\end{align}\]
Now, we will take out the term $2m$ common,
\[\begin{align}
  & \Rightarrow 2{{m}^{2}}-4m+m-2=0 \\
 & \Rightarrow 2m\left( m-2 \right)+1\left( m-2 \right)=0 \\
\end{align}\]
Again let’s take out $\left( m-2 \right)$ common, and then we will get: $\left( m-2 \right)\left( 2m-1 \right)=0$ , now to satisfy this equation either $\left( m-2 \right)=0\text{ or }\left( 2m+1 \right)=0$
Therefore the value of $m=2,\dfrac{-1}{2}$ , Now we will re-substitute $m$ as $\sin \theta $
Therefore $\sin \theta =2\text{ or }\dfrac{-1}{2}$ ,
Now since $\theta \in \left[ -2\pi ,2\pi \right]$$\Rightarrow \sin \theta =\dfrac{-1}{2}$
$\theta ={{\sin }^{-1}}\left( \dfrac{-1}{2} \right)\Rightarrow \theta =\left( 2\pi -\dfrac{\pi }{6} \right),\left( -\pi +\dfrac{\pi }{6} \right),\left( \pi +\dfrac{\pi }{6} \right),\dfrac{\pi }{6}$
Now we will find out the sum of the elements $S$ that is $\left( 2\pi -\dfrac{\pi }{6} \right),\left( -\pi +\dfrac{\pi }{6} \right),\left( \pi +\dfrac{\pi }{6} \right),\dfrac{-\pi }{6}$
$\Rightarrow \left( 2\pi -\dfrac{\pi }{6} \right)+\left( -\pi +\dfrac{\pi }{6} \right)+\left( \pi +\dfrac{\pi }{6} \right)+\left( -\dfrac{\pi }{6} \right)=2\pi $
Therefore the answer is Option A.

Note:Students can ignore the substitution part and can proceed with solving the equation in $\sin \theta $. At the end remember that the range of $\theta $ is $\left[ -2\pi ,2\pi \right]$so avoid all the values outside this range. This is the reason we have ignored the other value of m=2.