# How do you solve sin(2x)cos(x)=sin(x)?

May 26, 2018

$x = n \pi , 2 n \pi \pm \left(\frac{\pi}{4}\right) , \mathmr{and} 2 n \pi \pm \left(\frac{3 \pi}{4}\right)$ where $n \in \mathbb{Z}$

#### Explanation:

$\rightarrow \sin 2 x \cos x = \sin x$

$\rightarrow 2 \sin x \cdot {\cos}^{2} x - \sin x = 0$

$\rightarrow \sin x \left(2 {\cos}^{2} x - 1\right) = 0$

$\rightarrow \rightarrow \sin x \cdot \left(\sqrt{2} \cos x + 1\right) \cdot \left(\sqrt{2} \cos x - 1\right) = 0$

When $\sin x = 0$

$\rightarrow x = n \pi$

When $\sqrt{2} \cos x + 1 = 0$

$\rightarrow \cos x = - \frac{1}{\sqrt{2}} = \cos \left(\frac{3 \pi}{4}\right)$

$\rightarrow x = 2 n \pi \pm \left(\frac{3 \pi}{4}\right)$

When $\sqrt{2} \cos x - 1 = 0$

$\rightarrow \cos x = \frac{1}{\sqrt{2}} = \cos \left(\frac{\pi}{4}\right)$

$\rightarrow x = 2 n \pi \pm \left(\frac{\pi}{4}\right)$

May 26, 2018

$x = n \pi , \frac{\pi}{4} + n \pi , \frac{3 \pi}{4} + n \pi$ where $n \in \mathbb{Z}$

#### Explanation:

We have,

$\textcolor{w h i t e}{\times x} \sin 2 x \cos x = \sin x$

$\Rightarrow 2 \sin x \cos x \times \cos x = \sin x$ [As, $\sin 2 x = 2 \sin x \cos x$]

$\Rightarrow 2 \sin x {\cos}^{2} x - \sin x = 0$

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

Now,

Either,

$\sin x = 0 \Rightarrow x = {\sin}^{-} 1 \left(0\right) = n \pi$, where $n \in \mathbb{Z}$

Or,

$\textcolor{w h i t e}{\times x} 2 {\cos}^{2} x - 1 = 0$

$\Rightarrow 2 {\cos}^{2} x - \left({\sin}^{2} x + {\cos}^{2} x\right) = 0$ [As ${\sin}^{2} x + {\cos}^{2} x = 1$]

$\Rightarrow 2 {\cos}^{2} x - {\sin}^{2} x - {\cos}^{2} x = 0$

$\Rightarrow {\cos}^{2} x - {\sin}^{2} x = 0$

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

So, Either $\cos x - \sin x = 0 \Rightarrow \cos x = \sin x \Rightarrow x = \frac{\pi}{4} \pm n \pi$, where $n \in \mathbb{Z}$

Or,

$\cos x + \sin x = 0 \Rightarrow \cos x = - \sin x \Rightarrow x = \frac{3 \pi}{4} \pm n \pi$, where $n \in \mathbb{Z}$

So, Summing it all up,

$x = n \pi , \frac{\pi}{4} \pm n \pi , \frac{3 \pi}{4} \pm n \pi$, where $n \in \mathbb{Z}$