The Solution set of the equation $\sin ^2 \theta-\cos \theta=\frac{1}{4}$ in the interval $[0,2 \pi]$ is

The Solution set of the equation $\sin ^2 \theta-\cos \theta=\frac{1}{4}$ in the interval $[0,2 \pi]$ is
  1. $\left\{\frac{\pi}{6}, \frac{5 \pi}{6}\right\}$
  2. $\left\{\frac{\pi}{3}, \frac{5 \pi}{3}\right\}$
  3. $\left\{\frac{\pi}{3}, \frac{2 \pi}{3}\right\}$
  4. $\left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}$

Solution

$\begin{aligned} & \sin ^2 \theta-\cos \theta=\frac{1}{4} \\ & \left(1-\cos ^2 \theta\right)-\cos \theta=\frac{1}{4} \\ & 4-4 \cos ^2 \theta-4 \cos \theta-1=0 \\ & \Rightarrow 4 \cos ^2 \theta+4 \cos \theta-3=0 \\ & \Rightarrow \cos \theta=\frac{-1}{2} \text { or } \cos \theta=\frac{1}{2} \\ \therefore \quad & \theta=2 \pi-\frac{\pi}{3}, \frac{\pi}{3} \\ \therefore \quad & \theta=\left\{\frac{5 \pi}{3}, \frac{\pi}{3}\right\}\end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 2)

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