Question

How do you solve the equation `sin^2x=3cos^2x`?

Answer 1

`x=pi/3,(2pi)/3,(4pi)/3,(5pi)/3`

Explanation:

`(sinx)^2=3(cosx)^2`
`(sinx)^2=3(1-(sinx)^2)`
`(sinx)^2=3-3(sinx)^2`
`4(sinx)^2=3`
`(sinx)^2=3/4`
`sinx=+-(sqrt3/2)`
`x=pi/3,pi-pi/3,pi+pi/3,(2pi)-pi/3`
`x=pi/3,(2pi)/3,(4pi)/3,(5pi)/3`

If this was in the region `0<=x<=2pi`