Question

How do you solve $\cos 2x=\sin x$ on the interval $0\le x\le 2\pi$?

Answer 1

Determine the value of $\mathit{x}$:

Use the cosine of a double angle identities, $\cos 2A=1-2{\sin }^{2}A=2{\cos }^{2}A-1$:

$$\begin{gathered} \cos 2x=\sin x \\ \Rightarrow 1-2{\sin }^{2}x=\sin x \\ \Rightarrow 2{\sin }^{2}x+\sin x-1=0 \\ \Rightarrow 2{\sin }^{2}x+2\sin x-\sin x-1=0 \\ \Rightarrow 2\sin x\left(\sin x+1\right)-1\left(\sin x+1\right)=0 \\ \Rightarrow \left(2\sin x-1\right)\left(\sin x+1\right)=0 \end{gathered}$$

Equate each factor to zero:

$$\begin{gathered} 2\sin x-1=0 \\ \Rightarrow \sin x=\frac{1}{2}\mathrm{for}\left(0\le x\le 2\mathrm{\pi }\right) \\ \Rightarrow x=\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6} \end{gathered}$$

And,

$$\begin{gathered} \sin x+1=0 \\ \Rightarrow \sin x=-1for\left(0\le x\le 3\mathrm{\pi }\right) \\ \Rightarrow x=\frac{3\mathrm{\pi }}{2} \end{gathered}$$

Hence, the value of $\mathit{x}$ are $\frac{\mathbf{\pi }}{\mathbf{6}}\mathbf{,}\frac{\mathbf{5}\mathbf{\pi }}{\mathbf{6}}$ and $\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{2}}$ respectively.