Find the general solution to $\sin(4x)-\cos(x)=0$ 

Question: Find the general solution to $\sin(4x)-\cos(x)=0$



My attempt:
$$ \sin(4x)-\cos(x)=0$$
$$ \Leftrightarrow \sin(4x) = \cos(x) $$
$$ \Leftrightarrow \sin(4x) = \sin( \frac{\pi}{2} -x)$$
From another post I learnt that you can equate $\sin(x) = \sin(y)$ on 2 conditions so applying it here:
$$ \Leftrightarrow 4x = \frac{\pi}{2} -x + 2 \pi n$$ 
and 
$$ \Leftrightarrow4x = \pi - (\frac{\pi}{2} -x)+ 2 \pi n$$
Solving both for $x$
$$ x = \frac{\pi}{10}  + \frac{2\pi n }{5}$$
$$ x = \frac{\pi}{6}  + \frac{2\pi n }{3}$$
However I checked Wolfram alpha and they have different solutions? 

Am I correct or not?

$\sin(4x)−\cos(x)=0$
$2\sin(2x)\cos(2x)-\cos(x)=0$
$4\sin(x)\cos(x)(1-2\sin^2(x))-\cos(x)=0$
One possible solution is $\cos(x)=0$
$4\sin(x)(1-2\sin^2(x))=1$
$8\sin^3(x)-4\sin(x)+1=0$
Now, let $\sin(x)=m$ and solve the resulting cubic... 
