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 $RHS &gt;0$ and $LHS &lt; 0$

And keep in mind that $\csc x$ is undefined at certain points.

Actually it is a little trickier than that.  Here is a picture.
your inequality can be simplified to
$$\frac{1}{2\sin(x)\cos(x)}\le \frac{1}{\frac{1}{2}\sqrt{3}\cos(x)-\frac{1}{2}\sin(x)}$$
substituting $$a=\sin(x),b=\cos(x)$$ we get
$$\frac{1}{ab}\le \frac{2}{\sqrt{3}b-a}$$ with $a^2+b^2=1$
this gives $$-\frac{a-\sqrt{3}b+4ab}{2ab(a-\sqrt{3}b)}\geq 0$$