Find the value of $x$ given that $\cos(3x)\tan(5x) = \sin(7x)$ 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 $x=n\pi$
 A: Use the product-to-sum identities:
$$\sin7x=\cos3x\tan5x=\frac{\cos3x\sin5x}{\cos5x}=\frac{\frac12\left(\sin8x-\sin(-2x)\right)}{\cos5 x}\;\implies$$
$$2\cos5x\sin7x=\sin8x+\sin2x\;\implies\sin12x+\sin2x=\sin8x+\sin2x\implies$$
$$\sin12x=\sin8x\implies 12x=\begin{cases}8x\implies x=n\pi\\{}\\\pi-8x+2n\pi\implies x=\cfrac\pi{20}+\cfrac n{10}\pi\end{cases}\;\;\;\;\;,\;\;\;\; n\in\Bbb Z$$
A: The quantity $\sin (2x)$ does indeed vanish when $x = n\pi/2$.  However, in the first step you multiplied by $\cos(5x)$, and $\cos(5x) = 0$ when $x = n \pi / 2$ for any odd value of $n$.  This means that these solutions may be spurious, and we have to check whether they actually work in our original equation.  In fact, it is easy to see that they are not solutions to the original equation, since $\tan(5 x)$ is undefined when $x = n \pi / 2$ & $n$ is odd.
This means that only the roots of the form $x = n \pi/2$ where $n$ is even are actually roots of the original equation.  But that is equivalent to simply looking at numbers of the form $x = n \pi$ instead.
[Also: note that the roots of $\cos(10x)$ are $x = (2n+1)\pi/20$, not $(2n+1)\pi/2$.]
