Question

Solve the following equation:$\cos 2x+3\sin x=2$

Answer 1

Solve $\cos 2x+3\sin x=2$

$$\begin{gathered} \cos 2x+3\sin x=2 \\ \begin{array}{l}\Rightarrow 1-2{\sin }^{2}x+3\sin x=2\left[\because \cos 2A=1-2{\sin }^{2}A\right]\\ \Rightarrow 2{\sin }^{2}x-3\sin x+1=0\\ \Rightarrow \left(2\sin x-1\right)\left(\sin x-1\right)=0\end{array} \end{gathered}$$

So, $2\sin x-1=0$ or $\sin x-1=0$

$$\begin{gathered} \mathrm{For},\mathrm{si}nx=\frac{1}{2} \\ \begin{array}{l}\sin x=\frac{1}{2}\\ \Rightarrow x={\sin }^{-1}\left(\frac{1}{2}\right)\\ \Rightarrow x=n\pi +{\left(-1\right)}^n\frac{\pi }{6}\end{array} \end{gathered}$$

And

$$\begin{gathered} \mathrm{For},\sin x=1 \\ \begin{array}{l}\sin x=1\\ \Rightarrow x={\sin }^{-1}1\\ \Rightarrow x=2n\pi +\frac{\pi }{6}\end{array} \end{gathered}$$

Hence the solution of the given equation $\cos 2x+3\sin x=2$ is either $x=n\pi +{\left(-1\right)}^n\frac{\pi }{6}$ or $x=2n\pi +\frac{\pi }{6}$.