Question

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

Answer 1

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.`