Question

4 cos2 ϴ = 3 find ϴ?

Answer 1

or is it a typo?

Answer 2

\[4cos(2ϴ)=3\]\[cos(2ϴ)=\frac{3}{4}\]\[2ϴ=2n\pi+cos^{-1}(\frac{3}{4})\]\[ϴ=n\pi+\frac{1}{2}cos^{-1}(\frac{3}{4})\] where n is an integer

Answer 3

OR
cos(2x) = 3/4
cos^2 - sin^2 = 3/4
cos^2 -(1-cos^2) = 3/4
2cos^2 -1 = 3/4
2cos^2 = 7/4
cos^2 = 7/8
cos(x) = +-sqrt(7)/2sqrt2

Answer 4

its more likely cos^2 rather than cos(2x) in this case

->cos^2 = 3/4
cos(x) = +-sqrt3/2

x = pi/6, 5pi/6

Answer 5

i think you're right @dumbcow i didn't even think about the cos^2 :)