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:
$$\frac{2\sin x\cos x}{2\cos^2x-1} = \frac{\sin x}{\cos x}$$
$$\frac{2\cos x}{2\cos^2x-1} = \frac{1}{\cos x}$$
$$2\cos{^2}(x) = 2\cos{^2}(x) - 1$$
$$0 = -1 ????$$
 A: 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, $$\tan(x) = 0 \implies x = n \pi$$
A: Hint:
$$\tan(2x) = \tan(x) \quad \Rightarrow \\
\frac{\sin(2x)}{\cos(2x)} = \frac{\sin(x)}{\cos(x)} \quad \Rightarrow \\
\frac{2\sin(x)\cos(x)}{2\cos^2(x) -1} = \frac{\sin(x)}{\cos(x)} \quad \Rightarrow \\
\frac{2\cos^2(x)}{2\cos^2(x) - 1} = 1 \quad \text{ or }\quad \sin(x) = 0.
$$
It looks like you got the "other solution $\sin(x) = 0$. Of course you can't have $0 = -1$. All that means is just that $2\cos^2(x)$ is never equal to $2\cos^2(x) = 1$. So that specific equation does not have any solution. But again, there is the possibility that $\sin(x)= 0$ (which is equivalent to $\tan(x) = 0$.
A: 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
]
