How do you find all the solutions in the interval [0, 2π) without using a calculator of #3sinx = 2cos^2x#?

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How do you find all the solutions in the interval [0, 2π) without using a calculator of #3sinx = 2cos^2x#?

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Answer 1 · 2015-11-24T05:01:26

Solve 3sin x = 2cos^2 x

Ans: #pi/6 and (5pi)/6#

Explanation:

#3sin x = 2cos^2 x#
#3sin x = 2( 1 - sin^2 x) = 2 - 2sin^2 x#
#2sin^2 x + 3sin x - 2 = 0#
Solve this quadratic equation in sin x.
#D = d^2 = b^2 - 4ac = 9 + 16 = 25# --> #d = +- 5#
#sin x = -3/4 +- 5/4#
2 solutions --> sin x = 1/2 and sin x = - 2 (rejected as < -1).
#sin x = 1/2# --> #x = pi/6# and #x = (5pi)/6# (unit circle)

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How do you find all the solutions in the interval [0, 2π) without using a calculator of #3sinx = 2cos^2x#?

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