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?

$$\dfrac\pi{12}+\pi k=\pi n-\dfrac{11\pi}{12}$$
$$\iff k=n-1$$
Now for odd $k,k=2m+1$(say)
$\dfrac\pi8+\dfrac{\pi k}2=\dfrac\pi8+\dfrac{\pi(2m+1)}2=m\pi+\dfrac{5\pi}8=(m+1)\pi-\dfrac{3\pi}8$
For even $k,k=2m$(say), $\dfrac\pi8+\dfrac{\pi k}2=m\pi+\dfrac\pi8=(m+1)\pi-\dfrac{7\pi}8$
So, there must be mistake in the W.A. unless there is some typo in your input
