Question

How do you solve `4sin^2x=1` for x in the interval [0,2pi)?

Answer 1

`S={pi/6, (5pi)/6, (7pi)/6, (11pi)/6}`

Explanation:

`sin^2x =1/4`

`sinx = +- 1/2`

`x=sin^-1(+- 1/2)`

`x=pi/6, (5pi)/6, (7pi)/6, (11pi)/6`

`S={pi/6, (5pi)/6, (7pi)/6, (11pi)/6}`