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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.

Arin Chaudhuri
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2 Answers2

1

$\cos x = \sin \frac \pi2 - x \implies \sec (x + \frac{\pi}{6} = \csc(\frac {\pi}{3} -x)\\ \csc 2x \le \csc (\frac \pi3 - x)$

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. enter image description hereHere is a picture.

Doug M
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  • Thanks for this tip ! –  Dec 29 '17 at 19:16
  • Actually I try to find the following solutions for hours : $2\pi n - \frac{2}{3}\pi < x \leq 2\pi n - \frac{5\pi}{9}, 2\pi n - \frac{\pi}{2} < x < 2\pi n, 2\pi n + \frac{\pi}{9} \leq x < 2\pi n + \frac{\pi}{3}, 2\pi n + \frac{\pi}{2} < x \leq 2\pi n + \frac{2\pi}{3}, 2\pi n + \frac{7\pi}{9} \leq x < 2\pi n + \pi$. –  Dec 29 '17 at 19:23
  • does the picture help? $x\in [\frac {\pi}{9} + 2k\pi,\frac {\pi}{3} + 2k\pi)\cup(\frac {7\pi}{9} + 2k\pi,\pi+ 2k\pi) \cup (\frac {4\pi}{3} + 2k\pi,\frac {13\pi}{9} + 2k\pi)\cup(\frac {3\pi}{2}+ 2k\pi, 2\pi + 2k\pi)$ – Doug M Dec 29 '17 at 19:29
  • For everything in Q4, $\csc 2x < 0$ and $\csc (\frac {\pi}{3} - x) > 0$ – Doug M Dec 29 '17 at 19:31
-1

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 $$-\frac{a-\sqrt{3}b+4ab}{2ab(a-\sqrt{3}b)}\geq 0$$

Dr. Sonnhard Graubner
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  • Should it not be $sec(x + \frac{\pi}{6}) = \frac{1}{\frac{1}{2} \sqrt{3} cos(x) - \frac{1}{2}sin(x)}$ ? –  Dec 29 '17 at 17:30
  • thank you very much for your hint! have a nice time! – Dr. Sonnhard Graubner Dec 29 '17 at 17:32
  • Thank you for your answer. Your method is great! I now try to continue this exercise. I think the rest is easily done :). –  Dec 29 '17 at 17:41
  • we Can also compare our answers together – Dr. Sonnhard Graubner Dec 30 '17 at 13:29
  • I would be very pleased if you could post your detailed solution. Your answer helped me a little bit further, but the rest of the exercise was not so easy for me than I thought in the beginning :(. In fact, this is the first trigonometric inequality that I have to solve. I have only solved trig equations so far. –  Dec 30 '17 at 20:43