
Trig 2
another two that i am stuck:
cos4x7cos2x=8 with interval [0,2pi)
and tan^2(theta) + Tan (theta) 12 with the same interval.
for this one i have so far:
tan=4 and tan=3 meaning they are not special right triangles...so i have to use the calculator...but my teacher said that theres a way to graph it and the points of intersection are the answers

$\displaystyle \cos(4x) = \cos^{2}(2x)  \sin^{2}(2x)$
$\displaystyle \cos(4x) = 2\cos^{2}(2x)  1$
$\displaystyle \cos(4x) = 1  2\sin^{2}(2x)$
One of these will help you.