1. ## Trigonometric Equations

The equation is: $3 tan 2x - 4 = 6$
and I can get this equation down to $\frac{6tanx}{(3-3tan^2x)} = \frac{10-tan^2x}{3-3tan^2x}$ but no further.

the other problem is i can't prove the identity: $\frac{sin2x + sin6x}{sin10x - sin2x} = \frac{cos2x}{cos6x}$ but i don't even know where to start

I am starting to panic as this is a take home test due tomorrow and I am really getting stuck. Thanks a lot!

2. Originally Posted by qwerty723519@hotmail.com
The equation is: $3 tan 2x - 4 = 6$
and I can get this equation down to $\frac{6tanx}{(3-3tan^2x)} = \frac{10-tan^2x}{3-3tan^2x}$ but no further.

the other problem is i can't prove the identity: $\frac{sin2x + sin6x}{sin10x - sin2x} = \frac{cos2x}{cos6x}$ but i don't even know where to start

I am starting to panic as this is a take home test due tomorrow and I am really getting stuck. Thanks a lot!
$3\tan(2x)-4=6\Rightarrow{3\tan(2x)=10}$

Dividing we get $\tan(2x)=\frac{10}{3}$

Taking the arctan of both sides we get

$2x=arctan\bigg(\frac{10}{3}\bigg)\Rightarrow{x=\fr ac{\arctan\bigg(\frac{10}{3}\bigg)}{2}}$

3. Originally Posted by qwerty723519@hotmail.com
The equation is: $3 tan 2x - 4 = 6$
and I can get this equation down to $\frac{6tanx}{(3-3tan^2x)} = \frac{10-tan^2x}{3-3tan^2x}$ but no further.

the other problem is i can't prove the identity: $\frac{sin2x + sin6x}{sin10x - sin2x} = \frac{cos2x}{cos6x}$ but i don't even know where to start

I am starting to panic as this is a take home test due tomorrow and I am really getting stuck. Thanks a lot!
For the second one try here

Prosthaphaeresis Formulas -- from Wolfram MathWorld

4. Thanks, I guess the first doesn't work out nicely. Regarding the second, I actually used those formulas for several of the other problems I had to do, but was unsure how to apply them in this case