0
$\begingroup$

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:

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

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?

thanks

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.