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