Question

How do you solve `tanxsin^2x=3/4tanx` in the interval `0<=x<=2pi`?

Answer 1

`tanxsin^2x - 3/4tanx = 0`

`tanx(sin^2x - 3/4) = 0`

`tanx= 0 and sinx = +-sqrt(3)/2`

`x = 0, pi, pi/3, (2pi)/3, (4pi)/3 and (5pi)/3`

Hopefully this helps!