Question

How do you find all the solutions between 0 and 2π for `4sin^2x-3 = 0`?

Answer 1

`x in {pi/3, (2pi)/3, (5pi)/3, (7pi)/3}`

Explanation:

If `4sin^2x-3=0`
then
`color(white)("XXX")sin^2x=3/4`
and
`color(white)("XXX")sin x=+-sqrt(3)/2`

This is one of the standard reference angles `= pi/3`

Between 0 and `2pi`, the possibilities for this reference angle are
`color(white)("XXX")pi/3,(2pi)/3, (5pi)/3, (7pi)/3`