<s>[INST] Can't find solutions for $\tan{2x} = \tan{x}$ Solving $\tan{2x}=\tan{x}$
Reducing the left side: $$\frac{\sin{2x}}{\cos{2x}} = \frac{2\sin{x}\cos{x}}{2\cos^{2}(x)-1}.$$
Reducing the right side: $$\tan{x} = \frac{\sin{x}}{\cos{x}}.$$
therefore:
$$\frac{2\sin x\cos x}{2\cos^2x-1} = \frac{\sin x}{\cos x}$$
$$\frac{2\cos x}{2\cos^2x-1} = \frac{1}{\cos x}$$
$$2\cos{^2}(x) = 2\cos{^2}(x) - 1$$
$$0 = -1 ????$$
 [/INST] We know $\tan A=\tan B\implies A=n\pi+B$ where $n$ is any integer (Proof Below )
So, $\tan2x=\tan x\implies 2x=n\pi+x\implies x=n\pi$ where $n$ is any integer
[
 Proof: 
$\tan A=\tan B\implies \sin A\cos B=\cos A\sin B$
$\implies \sin(A-B)=0\implies A-B=n\pi$  where $n$ is any integer
]
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