Question

Solve,

2 sin 5x cos 3x - sin 4x = 0

Answer 1

$\begin{aligned}&2\sin(5x)\cos(3x)-\sin(4x)=0\\\\\implies\ \ &2\cdot\dfrac{1}{2}\left[\sin(5x+3x)+\sin(5x-3x)\right]-\sin(4x)=0\\\\\implies\ \ &\sin(8x)+\sin(2x)-\sin(4x)=0\\\\\implies\ \ &\sin(8x)-\sin(4x)+\sin(2x)=0\\\\\implies\ \ &2\cos\left(\dfrac{8x+4x}{2}\right)\sin\left(\dfrac{8x-4x}{2}\right)+\sin(2x)=0\\\\\implies\ \ &2\cos(6x)\sin(2x)+\sin(2x)=0\\\\\implies\ \ &\sin(2x)[2\cos(6x)+1]=0\end{aligned}$

$\begin{aligned}\\\\\implies\ \ &\sin(2x)=0\quad\quad\ \quad\longrightarrow\quad(1)\quad\\\\\&\ \ &2\cos(6x)+1=0\quad\longrightarrow\quad(2)\end{aligned}$

$\begin{aligned}(1)\implies\ \ &2x=n\pi\\\\\implies\ \ &x=\dfrac{n\pi}{2},\quad n\in\mathbb{Z}\\\\\\(2)\implies\ \ &\cos(6x)=-\dfrac{1}{2}\\\\\implies\ \ &\cos(6x)=\cos\left(\dfrac{2\pi}{3}\right)\\\\\implies\ \ &6x=2n\pi\pm\dfrac{2\pi}{3}\\\\\implies\ \ &x=\dfrac{n\pi}{3}\pm\dfrac{\pi}{9}\end{aligned}$