How do you solve #2cos^2x + 3cos x - 2 = 0 # on the interval #[0,2pi]#?

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How do you solve #2cos^2x + 3cos x - 2 = 0 # on the interval #[0,2pi]#?

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Answer 1 · 2015-07-25T23:26:47

Solve #f(x) = 2cos^2 x + 3cos x - 2 = 0.#

Ans: #+- pi/3#

Explanation:

Call cos x = t. Solve the quadratic equation:
#2t^2 + 3t - 2 = 0#
#D = d^2 = 9 + 16 = 25# --> #d = +- 5#
#t = cos x = -3/4 +- 5/4#
#t = 1/2# and t = -2 (rejected)
Trig table gives --> #cos x = 1/2# --> #x = +- pi/3#

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How do you solve #2cos^2x + 3cos x - 2 = 0 # on the interval #[0,2pi]#?

1 Answer

Explanation:

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