SOLUTION: Find the exact solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list.)
cos 2x + cos x = 0
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-> SOLUTION: Find the exact solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list.)
cos 2x + cos x = 0
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Question 1055398: Find the exact solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list.)
cos 2x + cos x = 0 Answer by ikleyn(52787) (Show Source):
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Find the exact solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list.)
cos 2x + cos x = 0
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cos(2x) + cos x = 0.
Replace cos(2x) by {{cos^2(x) - sin^2(x)}}} = = .
Then you get
= 0.
Factor left side:
(2cos(x) - 1)*(cos(x) +1) = 0.
This equation deploys in two independent equations:
1. 2cos(x) - 1 = 0 ---> cos(x) = ---> x = and x = in the given interval, and
2. cos(x) + 1 = 0 ---> cos(x) = -1 ---> x = in the given interval.
Answer. The solutions are , and .
