How do you solve 2sin^2 x-sinx-1=0?

1 Answer
Nghi N.
Jul 20, 2015

Solve 2sin^2 x - sin x - 1 = 0

Answer: pi/2; (7pi)/6 and (11pi)/6

Explanation:

Call sin x = t.
2t^2 - t - 1 = 0.
Since a + b + c = 0. use shortcut. 2 real roots: t = 1 and t = c/a = -1/2
sin x = t = 1 --> x = pi/2
sin x = t = -1/2 --> x = (7pi)/6 and x = (11pi)/6

Within interval (0, 2pi): 3 answers: pi/2; (7pi)/6 , and (11pi)/6.

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