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**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.