Question

Solve $2co{s}^{2}x+3sinx=0$.

Answer 1

We have given:

$2co{s}^{2}x+3sinx=0$

$$\begin{gathered} \Rightarrow 2(1-{\sin }^{2}x)+3\sin x=0[\because {\cos }^{2}x=1-{\sin }^{2}x] \\ \Rightarrow 2-2{\sin }^{2}x+3\sin x=0 \\ \Rightarrow 2{\sin }^{2}x-3\sin x-2=0 \\ \Rightarrow 2{\sin }^{2}x-4\sin x+\sin x-2=0 \\ \Rightarrow 2\sin x(\sin x-2)+1(\sin x-2)=0 \\ \Rightarrow (\sin x-2)(2\sin x+1)=0 \\ \text{Either}\sin x-2=0\text{or}2\sin x+1=0 \\ \Rightarrow \sin x=2\text{or}\sin x=\frac{-1}{2} \end{gathered}$$

Since, $\sin x=2$ is not possible as the range of $\sin x$ is $[-1,1]$.

So, for $\sin x=\frac{-1}{2}$, we know that $\sin$ is negative in third and fourth quadrant so.

So, $$\begin{gathered} x=\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\mathrm{or}\mathrm{x}=2\mathrm{\pi }-\frac{\mathrm{\pi }}{6}[\because \sin \frac{\mathrm{\pi }}{6}=\frac{1}{2}] \\ \Rightarrow \mathrm{x}=\frac{6\mathrm{\pi }+\mathrm{\pi }}{6}\mathrm{or}\mathrm{x}=\frac{12\mathrm{\pi }-\mathrm{\pi }}{6} \\ \Rightarrow \mathrm{x}=\frac{7\mathrm{\pi }}{6}\mathrm{or}\mathrm{x}=\frac{11\mathrm{\pi }}{6} \end{gathered}$$

Hence, $\mathbf{x}\mathbf{}\mathbf{=}\frac{\mathbf{7}\mathbf{\pi }}{\mathbf{6}}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathbf{x}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{11}\mathbf{\pi }}{\mathbf{6}}$ is the solution of given trigonometric function.