The general solution of the equation $3 \sec ^2 \theta=2 \operatorname{cosec} \theta$ is

The general solution of the equation $3 \sec ^2 \theta=2 \operatorname{cosec} \theta$ is
  1. $\mathrm{n} \pi+\frac{\pi}{4}, \mathrm{n} \in \mathrm{Z}$
  2. $2 \mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{12}, \mathrm{n} \in \mathrm{Z}$
  3. $\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{6}, \mathrm{n} \in \mathrm{Z}$
  4. $\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{3}, \mathrm{n} \in \mathrm{Z}$

Solution

$\begin{aligned} & 3 \sec ^2 \theta=2 \operatorname{cosec} \theta \\ & \Rightarrow \frac{3}{\cos ^2 \theta}=\frac{2}{\sin \theta} \\ & \Rightarrow \frac{3}{1-\sin ^2 \theta}=\frac{2}{\sin \theta} \\ & \Rightarrow 2 \sin ^2 \theta+3 \sin \theta-2=0 \\ & \Rightarrow(2 \sin \theta-1)(\sin \theta+2)=0 \\ & \Rightarrow \sin \theta=\frac{1}{2} \end{aligned}$ or $\sin \theta=-2$, which is not possible $\begin{aligned} \therefore \quad \sin \theta & =\frac{1}{2}=\sin \frac{\pi}{6} \\ \Rightarrow \theta & =\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{6}, \mathrm{n} \in Z \end{aligned}$

Asked in: MHT CET 2023 (13 May Shift 1)

Practice more Trigonometric Functions questions on Aicharya