Solution of $\sqrt{5-2\sin x}\geq 6\sin x-1$ 
Solve the following inequality. $$\sqrt{5-2\sin x}\geq 6\sin x-1.$$

My tries:
As $5-2\sin x>0$ hence we do not need to worry about the domain.
Case-1: $6\sin x-1\leq0\implies \sin x\leq\dfrac{1}{6}\implies -1\leq\sin x\leq\dfrac{1}{6}\tag*{}$
Case-2:$6\sin x-1>0\implies \dfrac{1}{6}<\sin x<1\tag*{}$
$\implies 5-2\sin x\geq36\sin^2x+1-12\sin x\implies 18\sin^2x-5\sin x-2\leq0$
$\implies(2\sin x-1)(9\sin x+2)\leq0$
$\implies\sin x\ \epsilon\ \bigg(\dfrac{1}{6},\dfrac{1}{2}\bigg]$
All of above implies $\sin x\ \epsilon\ \bigg[-1,\dfrac{1}{2}\bigg]$.
Answer is given in the form: $\bigg[\dfrac{\pi(12n-7)}{6},\dfrac{\pi(12n+1)}{6}\bigg]\ (n\epsilon Z)$
How do I reach the form given in options? I even don't know what I've is correct or not.
Please help.

Hint:)
Other way that may be helpful. Let $y=5-2\sin x$, from $\sqrt{5-2\sin x}\geq 6\sin x-1$ you find
$$3y-\sqrt{y}-14\geq0$$
gives
$(\sqrt{y}-2)(\sqrt{y}+\dfrac73)\geq0$. You can discuss about conditions and find that $\sin x\leq\dfrac12$. This concludes 
$$\color{blue}{\left[2k\pi+\dfrac{3\pi}{2}-\dfrac{2\pi}{3},2k\pi+\dfrac{3\pi}{2}+\dfrac{2\pi}{3}\right]}$$
