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

1 Answer
Nghi N.
Nov 24, 2015

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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