# Thread: Solving a trig equation

1. ## Solving a trig equation

Find exact values for all solutions to the following equation on the interval
0 ≤ t ≤ 2pi

2 cos 2t = 1 - 4 cos t
2(2cos^2 -1) = 1 - 4 cos t
4 cos ^2 t -2 = 1 - 4 cost t
4 cos^2 t + 4 cos t - 3 = 0
(2cos t - 1)(2cos t + 3) = 0

2cos t - 1 = 0
2cos t = 1
cos t = 1/2
t = pi/3, 5pi/3

2cos t = -3
cos t = -3/2
t = ?????

So for this problem, would we only have two solutions? It seems as though there are no solutions for the inverse of cos -3/2. Please confirm this. Thank you so much.

2. Originally Posted by florx
Find exact values for all solutions to the following equation on the interval
0 ≤ t ≤ 2pi

2 cos 2t = 1 - 4 cos t
2(2cos^2 -1) = 1 - 4 cos t
4 cos ^2 t -2 = 1 - 4 cost t
4 cos^2 t + 4 cos t - 3 = 0
(2cos t - 1)(2cos t + 3) = 0

2cos t - 1 = 0
2cos t = 1
cos t = 1/2
t = pi/3, 5pi/3

2cos t = -3
cos t = -3/2
t = ?????

So for this problem, would we only have two solutions? It seems as though there are no solutions for the inverse of cos -3/2. Please confirm this. Thank you so much.
Dear florx,

You are correct. Actually $\displaystyle 2cost-3\neq{0}$ since $\displaystyle cost\neq{\frac{-3}{2}}$