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:
$(x+15)=(90-2x)+k.360$ and
$(x+15)=-(90-2x)+k.360$
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
use that $$1+\tan^2(x)=\frac{\sin^2(x)+\cos^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}$$