The solution set of $8 \cos ^2 \theta+14 \cos \theta+5=0$, in the interval $[0,2 \pi]$, is

The solution set of $8 \cos ^2 \theta+14 \cos \theta+5=0$, in the interval $[0,2 \pi]$, is
  1. $\left\{\frac{\pi}{3}, \frac{2 \pi}{3}\right\}$
  2. $\left\{\frac{\pi}{3}, \frac{4 \pi}{3}\right\}$
  3. $\left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}$
  4. $\left\{\frac{2 \pi}{3}, \frac{5 \pi}{3}\right\}$

Solution

$\begin{array}{ll} & 8 \cos ^2 \theta+14 \cos \theta+5=0 \\ \therefore \quad & 8 \cos ^2 \theta+10 \cos \theta+4 \cos \theta+5=0 \\ \therefore \quad & 2 \cos \theta(4 \cos \theta+5)+1(4 \cos \theta+5)=0 \\ \therefore \quad & (2 \cos \theta+1)(4 \cos \theta+5)=0 \\ \therefore \quad & \cos \theta=\frac{-1}{2} \text { or } \cos \theta=\frac{-5}{4} \end{array}$ But $\cos \theta=\frac{-5}{4}$ is not possible as $\cos \theta \in[-1,1]$ for all values of $\theta$. $\begin{aligned} & \therefore \quad \cos \theta=\frac{-1}{2} \\ & \therefore \quad \theta \in\left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\} \end{aligned}$

Asked in: MHT CET 2023 (12 May Shift 1)

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