# Math Help - Trig question

1. ## Trig question

Solve the equation sin2A -cos^2A=0
giving values of A between 0 degrees and 360 degrees.

THANKS!

2. Originally Posted by Oasis1993
Solve the equation sin2A -cos^2A=0
giving values of A between 0 degrees and 360 degrees.

THANKS!
Substitute sin(2A) = 2 sin(A) cos(A). Then factorise and use the null factor law to get two simple trig equatins to solve.

3. I didnt understand. How am i supposed to factorise it?

4. Originally Posted by Oasis1993
Solve the equation sin2A -cos^2A=0

Originally Posted by Oasis1993

giving values of A between 0 degrees and 360 degrees.

THANKS!

hi all
my solution

5. Originally Posted by Oasis1993
I didnt understand. How am i supposed to factorise it?
Did you substitute sin(2A) = 2 sin(A) cos(A) like I said to? If so, you should see how to factorise by taking out the common factor of cos(A).