Question

Find the value of x:$\cot 4x(\sin 5x+\sin 3x)=\cot x(\sin 5x-\sin 3x)$

Answer 1

$\cot 4x(\sin 5x+\sin 3x)=\cot x(\sin 5x-\sin 3x)$

L.H.S$=\cot 4x\left[2\sin \left(\frac{5x+3x}{2}\right)\cos \left(\frac{5x-3x}{2}\right)\right]$ using transformation angle formula $\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$

$=2\cot 4x\sin 4x\cos x$ .....$\left(1\right)$

R.H.S$=\cot x(\sin 5x-\sin 3x)$

$=\cot x\left[2\sin \left(\frac{5x-3x}{2}\right)\cos \left(\frac{5x+3x}{2}\right)\right]$ using transformation angle formula $\sin C-\sin D=2\sin \left(\frac{C-D}{2}\right)\cos \left(\frac{C+D}{2}\right)$

$=2\cot x\sin x\cos 4x$ .....$\left(2\right)$

Equating $\left(1\right)$ and $\left(2\right)$ we have

$\Rightarrow 2\cot 4x\sin 4x\cos x=2\cot x\sin x\cos 4x$

$\Rightarrow \frac{2\cot 4x\sin 4x\cos x}{2\cot x\sin x\cos 4x}=1$

$\Rightarrow \cot 4x\times \frac{\sin 4x}{\cos 4x}\times \frac{\cos x}{\sin x}=1$

$\Rightarrow \cos 4x\times \tan 4x\times \cos x=1$

$\therefore x=n\pi +\frac{\pi }{4}$