Solution of $\sqrt{5-2\sin x}\geq 6\sin x-1$

Question

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.

Answer 1

solving the inequality $$18\sin(x)^2-5\sin(x)-2\le 0$$ we have $$2 \pi c_1-\sin ^{-1}\left(\frac{2}{9}\right)\leq x\leq \frac{1}{6} \left(12 \pi c_1+\pi \right)\lor \frac{1}{6} \left(12 \pi c_1+5 \pi \right)\leq x\leq 2 \pi c_1+\pi +\sin ^{-1}\left(\frac{2}{9}\right)$$

Answer 2

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.

Answer 3

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 $$\color{blue}{\left[2k\pi+\dfrac{3\pi}{2}-\dfrac{2\pi}{3},2k\pi+\dfrac{3\pi}{2}+\dfrac{2\pi}{3}\right]}$$