Question

cos^2-sin^2= (sqrt 3/2) need help guys, please

Answer 1

well cos(2x) = cos^(x) - sin^2(x)

so then

\[\cos(2x) = \frac{\sqrt{3}}{2}\]

so use an exact value triangle to find 2x, then halve it to find x...

hope it helps

Answer 2

oops should read

\[\cos(2x) = \cos^2(x) - \sin^2(x)\]