Question

How do you solve `2sin^2x=sinx+1` for `0<=x<=2pi`?

Answer 1

`x = pi/2, 7/6pi and 11/6pi`

Explanation:

`2sin^2x = sin x+1`
`2sin^x - sin x - 1 =0`
`(2 sinx +1)(sinx-1) = 0`

`a. 2sinx + 1 =0, sinx = -1/2`
`x = sin^-1(-1/2)`
`x = 7/6(pi),11/6(pi)`

`b. sin-1 =0, sinx = 1`
`x = sin^-1(1)`
`x = pi/2`