# Trig 2

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• May 6th 2008, 06:09 PM
>_<SHY_GUY>_<
Trig 2
another two that i am stuck:

cos4x-7cos2x=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
• May 6th 2008, 07:38 PM
TKHunny
$\cos(4x) = \cos^{2}(2x) - \sin^{2}(2x)$

$\cos(4x) = 2\cos^{2}(2x) - 1$

$\cos(4x) = 1 - 2\sin^{2}(2x)$

One of these will help you.