Hi, I'm working on this problem, and it's throwing me off a bit because it's a little different from what I've been doing.
Solve 4(cos^2(3x)) − 3 = 0 for [0, 2π). Give exact answers.
Here is what I've done so far:
4 (cos^2(3x)) - 3 = 0
4 (cos^2(3x)) = 3 (added 3)
cos^2(3x) = 3/4 (divided by 4)
The three is throwing me off a little and I don't know what the next step is.
Please let me know if I can give anymore information or make anything clear.
Thank you, Dan.
I determined:
4 (cos^2(3x)) - 3 = 0
4 (cos^2(3x)) = 3 (added 3)
cos^2(3x) = 3/4 (divided by 4)
√cos^2(3x) = √3/4
cos3x = +/- √3/(2)
3x = π/6 + (nπ)
and
3x = 5π/6 + nπ
Final answer (dividing by 3)
x = π/18 + (nπ)/3
and
x = (5π)/18 + (nπ)/3