Question 17.2.10: Solve: 2cos² x+3sin x=0 ....
Solve: 2 \cos ^{2} x+3 \sin x=0 .
Step-by-Step
We have
\begin{aligned}& 2 \cos ^{2} x+3 \sin x=0\\ \\\Rightarrow & 2\left(1-\sin ^{2} x\right)+3 \sin x=0 \\ \\\Rightarrow & 2 \sin ^{2} x-3 \sin x-2=0\\ \\\Rightarrow & 2 \sin ^{2} x-4 \sin x+\sin x-2=0 \\ \\\Rightarrow & 2 \sin x(\sin x-2)+(\sin x-2)=0 \\ \\\Rightarrow & (\sin x-2)(2 \sin x+1)=0\\ \\\Rightarrow & (\sin x-2)=0 \text { or }(2 \sin x+1)=0 \\ \\\Rightarrow & \sin x=2 \text { or }(2 \sin x+1)=0\\ \\\Rightarrow & 2 \sin x+1=0 \quad[\because \sin x=2 \text { is not possible }]\\ \\\Rightarrow & \sin x=-\frac{1}{2}=-\sin \frac{\pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right)=\sin \frac{7 \pi}{6}\\ \\\Rightarrow & \sin x=\sin \frac{7 \pi}{6} \\ \\\Rightarrow & x=\left\{n \pi+(-1)^{n} \cdot \frac{7 \pi}{6}\right\}, \text { where } n \in I .\end{aligned}Hence, the general solution is given by x=\left\{n \pi+(-1)^{n} \cdot \frac{7 \pi}{6}\right\} , where n \in I .
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