I want to ask about trigonometric equation.

My effort:

First I think I need to change csc to sin and tan to sec and cos form.

$\dfrac {1}{\sin 2x}= {\sqrt {{\sec^2(x+15)}}}$

$\dfrac {1}{2\sin x\cos x}=\sqrt {\dfrac{1}{\cos^2(x+15)}}$

$\dfrac{1}{2\sin x\cos x}={\dfrac{1}{\cos(x+15)}}$

$\cos(x+15)=2\sin x\cos x$

since $\sin x= \cos(90-x)$, we can change it to:

$\cos(x+15)=2\cos(90-x)\cos x$

Then with the formula for cosine multiplication, we can get:


From there, we can apply the formula for cosine equation:

  1. $(x+15)=(90-2x)+k.360$ and

  2. $(x+15)=-(90-2x)+k.360$

Apply k=0,1,2 to both equationa and we got the solutions should be:

$25,105,$ and $145$

But, when I check online calculator, the solutions should be more than 3. Here's the link: http://www.wolframalpha.com/input/?i=csc+2x+%3D+sqrt(1%2Btan%5E2(x%2B15)),+0%3C%3Dx%3C%3D180

Can someone try to fix my mistakes and also provides the correct answer to me?


  • $\begingroup$ Do you mean $$\csc(2x)=\sqrt{1+\tan^2(x+15^{\circ})}$$? $\endgroup$ – Dr. Sonnhard Graubner Aug 17 '18 at 12:25
  • $\begingroup$ yes, just edit the title now, thanks for the correction $\endgroup$ – akusaja Aug 17 '18 at 12:26

use that $$1+\tan^2(x)=\frac{\sin^2(x)+\cos^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}$$

  • $\begingroup$ then, from here, how can I arrive at the solution? $\endgroup$ – akusaja Aug 17 '18 at 12:47
  • $\begingroup$ Use that $$\cos(x+15^{\circ})=\cos(x)\cos(15^{\circ})-\sin(x)\sin(15^{\circ})$$ $\endgroup$ – Dr. Sonnhard Graubner Aug 17 '18 at 12:53

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