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Question: Let \(S=\left[ \theta \in \left[ -2\pi ,2\pi \right]:2{{\cos }^{2}}\theta +3\sin \theta =0 \right]\)...

Let S=[θ∈[−2π,2π]:2cos⁡2θ+3sin⁡θ=0]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 SS is?
A. 13π6\dfrac{13\pi }{6}
B. 00
C. 2π2\pi
D. 5π3\dfrac{5\pi }{3}

Explanation

Solution

We will first convert the given equation in quadratic form, we will do this by writing cos⁡2θ{{\cos }^{2}}\theta as (1−sin⁡2θ)(1-{{\sin }^{2}}\theta ), we will then substitute sin⁡θ\sin \theta as mm, 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⁡θ\sin \theta and then find the inverse in order to find out the elements of SS.

Complete step-by-step answer :
We are given the range of θ\theta as [−2π,2π]\left[ -2\pi ,2\pi \right] and 2cos⁡2θ+3sin⁡θ=02{{\cos }^{2}}\theta +3\sin \theta =0, We will start by solving for θ\theta :
We have with us: 2cos⁡2θ+3sin⁡θ=0 ........ Equation 1.2{{\cos }^{2}}\theta +3\sin \theta =0\text{ }........\text{ Equation 1}\text{.}
Now we know that cos⁡2θ+sin⁡2θ=1⇒cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta
Now putting this value in our equation no. 1

& 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{aligned}$$ Taking the elements of L.H.S. to R.H.S. we will get the following equation: $$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{aligned} & \Rightarrow 2{{m}^{2}}-3m-2=0 \\\ & \Rightarrow 2{{m}^{2}}-4m+m-2=0 \\\ \end{aligned}$$ Now, we will take out the term $2m$ common, $$\begin{aligned} & \Rightarrow 2{{m}^{2}}-4m+m-2=0 \\\ & \Rightarrow 2m\left( m-2 \right)+1\left( m-2 \right)=0 \\\ \end{aligned}$$ 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.