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:
$$\frac{2\sin x\cos x}{2\cos^2x-1} = \frac{\sin x}{\cos x}$$
$$\frac{2\cos x}{2\cos^2x-1} = \frac{1}{\cos x}$$
$$2\cos{^2}(x) = 2\cos{^2}(x) - 1$$
$$0 = -1 ????$$

Recall that
$$\tan(A+B) = \dfrac{\tan(A) + \tan(B)}{1-\tan(A) \tan(B)}$$
Hence,
$$\tan(2x) = \dfrac{2\tan(x)}{1-\tan^2(x)} = \tan(x)$$
This implies $\tan(x) = 0 \text{ or }1-\tan^2(x)  = 2$. $1-\tan^2(x) = 2$ is not possible if $x \in \mathbb{R}$. Hence, $$\tan(x) = 0 \implies x = n \pi$$
