Just Another Regular Trig Problem

$\begin{aligned} \frac {\sin 5x}{\sin x} - \frac {\cos 5x}{\cos x} & = 2 \\ \frac {\sin 5x\cos x - \cos 5x \sin x}{\sin x \cos x} & = 2 \\ \sin 5x\cos x - \cos 5x \sin x & = 2 \sin x \cos x \\ \sin 4x & = \sin 2x \\ 2\sin 2x \cos 2x & = \sin 2x \\ \sin 2x (2\cos 2x - 1) & = 0 \\ \implies \cos 2x & = \frac 12 & \small \color{#3D99F6} \text{When }\sin 2x = 0 \text{, the equation is undefined.} \\ \implies x & = \frac \pi 6, \frac {5\pi}6, \frac {7\pi}6, \frac {11\pi}6 & \small \color{#3D99F6} \text{for }0 \le x < 2\pi \end{aligned}$

Therefore, there are $\boxed{4}$ solutions.

Cross Multiply:

$2=\frac { \sin { 5x } \cos { x } -\sin { x } \cos { 5x } }{ \sin { x } \cos { x } }$

Using the double angle formula:

$\boxed {\sin { (a-b)= } \sin { a } \cos { b } -\cos { a } \sin { b } }$

$a=5x,b=x$

$2=\frac { \sin { (5x-x) } }{ \sin { x } \cos { x } }$

$2=\frac { \sin { 4x } }{ \sin { x } \cos { x } }$

$\boxed { a=-b\Rightarrow \sin { (2a)= } 2\sin { a } \cos { a } }$

$2=\frac { 2\sin { 2x\cos { 2x } } }{ \sin { x } \cos { x } }$

$2=\frac { 2(2\sin { x } \cos { x } )\cos { 2x } }{ \sin { x } \cos { x } }$

$2=4\cos { 2x }$

$\frac { 1 }{ 2 } =\cos { 2x }$

Which has 4 solutions.