Asked by lance lazenby
Solve 2cos2𝑥+cos𝑥−1=0for all solutions in the interval [0,360°)
Answers
Answered by
oobleck
2cos2x + cosx - 1 = 0
2(2cos^2x-1) + cosx - 1 = 0
4cos^2x + cosx - 3 = 0
(4cosx-3)(cosx+1) = 0
cosx = 3/4 or -1
since cos41.4° = 3/4,
x = 41.4°, 180°, 360-41.4°
2(2cos^2x-1) + cosx - 1 = 0
4cos^2x + cosx - 3 = 0
(4cosx-3)(cosx+1) = 0
cosx = 3/4 or -1
since cos41.4° = 3/4,
x = 41.4°, 180°, 360-41.4°
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