Solve the trigonometric equation: $\cos (3x)-\sin(x)=\sqrt 3(\cos (x)-\sin(3x))$ 
Solve the trigonometric equation:

$$\cos (3x)-\sin(x)=\sqrt 3(\cos (x)-\sin(3x))$$

My answer is contradictory to Wolfram Alpha.
Because, W.A. gives me:

$x = \pi n - \frac {11 \pi}{12},   n \in \mathbb{ Z}$
$x = \pi n - \frac {7 \pi}{12},   n \in \mathbb{ Z}$
$x = \pi n - \frac {3 \pi}{12},   n \in \mathbb{ Z}$

But, my answer is:

$x=\frac {\pi}{12}+\pi k, k\in\mathbb{Z}$
$x=\frac {\pi}{8}+\frac {\pi k}{2}, k\in\mathbb{Z}$

Is my solution wrong? Or What is the problem in my solution?

$$\frac{\cos (3x)-\sin(x)}{\cos (x)-\sin(3x)}=\sqrt 3$$
$$\frac{\sin (\pi/2-3x)-\sin(x)}{\sin (\pi/2-x)-\sin(3x)}=\sqrt 3$$
$$\frac{2\cos(\pi/4-x)\sin(\pi/4-2x)}{2\cos(\pi/4+x)\sin(\pi/4-2x)}=\sqrt 3$$
$$\frac{\cos(\pi/4-x)}{\cos(\pi/4+x)}=\sqrt 3$$
$$\frac{\sin(\pi/2-\pi/4+x)}{\cos(\pi/4+x)}=\sqrt 3$$
$$\tan(\pi/4+x)=\tan(\pi/3)$$
then
$$\pi/4+x=k\pi+\pi/3$$
or
$$x=k\pi+\pi/12$$
