I converted $\tan(5x)$ into ${\sin(5x)/\cos(5x)}$ and multiplied $\cos(5x)$ to RHS resulting in $\cos(3x)\sin(5x) = \sin(7x)\cos(5x)$
Multiplying both sides by $2$ and using the formula for $2\sin(A)\cos (B)$ I got $\sin(8x) + \sin(2x) = \sin(12x) + \sin(2x)$
which turns into $\sin(12x) – \sin(8x) = 0$
Then using the formula for $\sin(C)-\sin(D)$ I got $2\cos (10x)\sin (2x) = 0$ which implies that either $\cos(10x)=0$ or $\sin(2x)=0$
After using the formulas for values I got $x= (2n+1)\pi/20$ or $x=n\pi/2$ but the answer is $x= (2n+1)\pi/20$ or $x=n\pi$
Best Answer
Use the product-to-sum identities:
$$\sin7x=\cos3x\tan5x=\frac{\cos3x\sin5x}{\cos5x}=\frac{\frac12\left(\sin8x-\sin(-2x)\right)}{\cos5 x}\;\implies$$
$$2\cos5x\sin7x=\sin8x+\sin2x\;\implies\sin12x+\sin2x=\sin8x+\sin2x\implies$$
$$\sin12x=\sin8x\implies 12x=\begin{cases}8x\implies x=n\pi\\{}\\\pi-8x+2n\pi\implies x=\cfrac\pi{20}+\cfrac n{10}\pi\end{cases}\;\;\;\;\;,\;\;\;\; n\in\Bbb Z$$