Solve for $x$ when $\sin 2x = \cos x$ where $ x$ is in the domain $ [0, 2\pi]$ Quick question on trig (which I haven't dealt with in a long time): 
since $\sin 2x = 2\sin x\cos x $
$2\sin x\cos x = \cos x$
$2\sin x\cos x/\cos x = 1$
$\sin x = 1/2$
since $\sin x = 1/2$ in quadrants $1$ and $2$, $x = \pi/6$ and $x = 5\pi/6$
Is this correct? If not, hint please.

Generalized Solution :
$$\sin ax=\cos bx$$ for example,  $\sin\dfrac{3x}2=\cos5x$ can also be addressed this way 
Method $\#1:$
$$\cos bx=\sin\left(\frac\pi2-bx\right)\implies\sin ax=\sin\left(\frac\pi2-bx\right)$$
$$\implies ax=n\pi+(-1)^n\left(\frac\pi2-bx\right)$$ where $n$ is any integer
If $n$ is odd, $=2m+1$(say), $ax=(2m+1)\pi-\left(\dfrac\pi2-bx\right)\implies x=\dfrac{(4m+1)\pi}{2(a-b)}$
If $n$ is even, $=2m$(say), $ x=\dfrac{(4m+1)\pi}{2(a+b)}$
Now we need to find the values of $m$ such that $0\le x\le2\pi$
Method $\#2:$
$$\sin ax=\cos\left(\frac\pi2-ax\right) \implies\cos bx=\cos\left(\frac\pi2-ax\right)$$
$$\implies bx=2r\pi\pm\left(\frac\pi2-ax\right)$$ where $r$ is any integer
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