Question

How do you solve `4sin^2x-1=0` in the interval `0<=x<=2pi`?

Answer 1

`pi/6, (5pi)/6, (7pi)/6 and (11pi)/6`

Explanation:

`4sin^2 x - 1 = 0`
`sin^2 x = 1/4`
`sin x = +- 1/2`
Use trig table of special arcs, and unit circle -->
a. When `sin x = 1/2` --> `x = pi/6 and x = (5pi)/6`
b. When `sin x = - 1/2` --> `x = (7pi)/6 and x = (11pi)/6`

Answers for `(0, 2pi)`
`pi/6, (5pi)/6, (7pi)/6, (11pi)/6`