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Question: If \(5\cos 2\theta +2{{\cos }^{2}}\dfrac{\theta }{2}+1=0\), where \(0<\theta <\pi \), then the value...

If 5cos⁡2θ+2cos⁡2θ2+1=05\cos 2\theta +2{{\cos }^{2}}\dfrac{\theta }{2}+1=0, where 0<θ<π0<\theta <\pi , then the value of θ\theta
A. π3±π\dfrac{\pi }{3}\pm \pi
B. π3,cos⁡−1(35)\dfrac{\pi }{3},{{\cos }^{-1}}\left( \dfrac{3}{5} \right)
C. cos⁡−1(35)±π{{\cos }^{-1}}\left( \dfrac{3}{5} \right)\pm \pi
D. π3,π−cos⁡−1(35)\dfrac{\pi }{3},\pi -{{\cos }^{-1}}\left( \dfrac{3}{5} \right)

Explanation

Solution

Hint : We use the multiple and submultiple angle formulas of cos⁡2θ=2cos⁡2θ−1\cos 2\theta =2{{\cos }^{2}}\theta -1 and 2cos⁡2θ2=1+cos⁡θ2{{\cos }^{2}}\dfrac{\theta }{2}=1+\cos \theta to simplify the given equation. We convert them to one quadratic equation and solve that to find the possible angles for the ratios. Not all the roots of the equation will be a solution , roots which are not in the range of for which the given trigonometric function doesn't exist will be discarded.

Complete step-by-step answer :
We first simplify the equation 5cos⁡2θ+2cos⁡2θ2+1=05\cos 2\theta +2{{\cos }^{2}}\dfrac{\theta }{2}+1=0 using the formulas of multiple and submultiple angles. We have
cos⁡2θ=2cos⁡2θ−1 2cos⁡2θ2=1+cos⁡θ \begin{aligned} & \cos 2\theta =2{{\cos }^{2}}\theta -1 \\\ & 2{{\cos }^{2}}\dfrac{\theta }{2}=1+\cos \theta \\\ \end{aligned}
The equation becomes
5cos⁡2θ+2cos⁡2θ2+1=0 ⇒5(2cos⁡2θ−1)+1+cos⁡θ+1=0 ⇒10cos⁡2θ+cos⁡θ−3=0 \begin{aligned} & 5\cos 2\theta +2{{\cos }^{2}}\dfrac{\theta }{2}+1=0 \\\ & \Rightarrow 5\left( 2{{\cos }^{2}}\theta -1 \right)+1+\cos \theta +1=0 \\\ & \Rightarrow 10{{\cos }^{2}}\theta +\cos \theta -3=0 \\\ \end{aligned}
We now simplify the quadratic and get

& 10{{\cos }^{2}}\theta +\cos \theta -3=0 \\\ & \Rightarrow 10{{\cos }^{2}}\theta +6\cos \theta -5\cos \theta -3=0 \\\ & \Rightarrow \left( 5\cos \theta +3 \right)\left( 2\cos \theta -1 \right)=0 \\\ \end{aligned}$$ So, values of $$\cos \theta $$ are $\cos \theta =-\dfrac{3}{5},\dfrac{1}{2}$. We know that in the principal domain or the periodic value of $0\le x\le \pi $ for $\cos x$, if we get $\cos x=\cos b$ where $0\le a,b\le \pi $ then $a=b$. We have $\cos \left( 126.87 \right)=-\dfrac{3}{5}$, $0\le {{126.87}^{\circ }}\le \pi $. We can also write it as ${{\cos }^{-1}}\left( \dfrac{3}{5} \right)$. We have $\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$, $0\le \dfrac{\pi }{3}\le \pi $. Therefore, $\cos \theta =-\dfrac{3}{5},\dfrac{1}{2}$ gives $\theta ={{\cos }^{-1}}\left( \dfrac{3}{5} \right),\dfrac{\pi }{3}$ as primary value. **So, the correct answer is “Option B”.** **Note** : Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $0\le x\le \pi $. In that case we have to use the formula $x=2n\pi \pm a$ for $\cos x=\cos b$ where $0\le x\le \pi $. For our given problem $\cos \theta =-\dfrac{3}{5},\dfrac{1}{2}$, the primary solution is $\theta ={{\cos }^{-1}}\left( \dfrac{3}{5} \right),\dfrac{\pi }{3}$. The general solution will be $x=\left( 2n\pi \pm {{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right)\cup \left( 2n\pi \pm \dfrac{\pi }{3} \right)$. Here $n\in \mathbb{Z}$.