Thread: Trig question

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

THANKS!

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.

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

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

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).