<s>[INST] 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.
 [/INST] Your solution is right.
You need only to write the answer, for which 
just take on the $y$-axis the point $\frac{1}{2}$ and you need $\sin{x}\leq\frac{1}{2}$, which gives $y\leq\frac{1}{2}$ and the arc
$$\left[-\frac{7\pi}{6},\frac{\pi}{6}\right]$$
on the trigonometric circle  and add $2\pi n$ in the both sides. 
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