How to solve the trigonometric inequality $\csc(2x) \leq \sec(x + \frac{\pi}{6})$ in $\mathbb{R}$? I have difficulty in solving the inequality 

$\csc(2x) \leq \sec(x + \frac{\pi}{6})$ in $\mathbb{R}$. 

First, I need to solve the equation $\csc(2x) - \sec(x + \frac{\pi}{6}) = 0$. I see that the period is $2\pi$. My problem is that I have no idea how to factor $\csc(2x) - \sec(x + \frac{\pi}{6})$. Which method can I use  to transform it into a product of basic trig equations? Or is there another way to solve this equation ?
Thanks for your help. 

$\cos x = \sin \frac \pi2 - x \implies \sec (x + \frac{\pi}{6} = \csc(\frac {\pi}{3} -x)\\
\csc 2x \le \csc (\frac \pi3 - x)$
When the LHS and RHS have the same sign: (careful, it is more involved than it appears on the face of it to find these regions.)
$2x \ge \frac \pi3 - x\\
x \ge \frac {\pi}{9} + k\frac {2\pi}{3}$
Plus the regions were the $RHS >0$ and $LHS < 0$
And keep in mind that $\csc x$ is undefined at certain points.
Actually it is a little trickier than that.  Here is a picture.
