Question

How do you solve $\sin \left(2x\right)=\sin \left(x\right)$ ?

Answer 1

Compute the required value.

Given: $\sin \left(2x\right)=\sin \left(x\right)$

Since, $\sin \left(2x\right)=2\sin \left(x\right)\cos \left(x\right)$

So, $2\sin \left(x\right)\cos \left(x\right)=\sin \left(x\right)$

$\Rightarrow$$2\sin \left(x\right)\cos \left(x\right)-\sin \left(x\right)=0$

$\Rightarrow$ $\sin \left(x\right)\left[2\cos \left(x\right)-1\right]=0$

Thus, $\sin \left(x\right)=0$ or $2\cos \left(x\right)-1=0$

If $\sin \left(x\right)=0$,

Since $\sin \left(\theta \right)$ is $0$, for $\theta =\mathrm{\pi }$ and $\sin x$ is periodic in $2\mathrm{\pi }$ period.

then $x=k\mathrm{\pi },\mathrm{k}=0,\pm 1,\pm 2.......$

If $2\cos \left(x\right)-1=0$

$\Rightarrow$ $\cos \left(x\right)=\frac{1}{2}$

Since $\cos \left(\theta \right)$ is $\frac{1}{2}$, for $\theta =\frac{\mathrm{\pi }}{3}$ and $\cos x$ is periodic in $2\mathrm{\pi }$ period.

then, $x=\frac{\mathrm{\pi }}{3}+2k\mathrm{\pi },\mathrm{k}=0,\pm 1,\pm 2........$

Hence the solutions in the interval $\left[0,2\mathrm{\pi }\right]$ are: $0,\mathrm{\pi },2\mathrm{\pi },\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}$