Question

How do you solve `4cotx=cotxsin^2x` for `0<=x<=2pi`?

Answer 1

`x=pi/2` and `x=(3pi)/2`

Explanation:

`4cotx=cotxsin^2x`

`hArrcotxsin^2x-4cotx=0`

or `cotx(sin^2x-4)=0`

As `sin^2x` ranges from `0` to `1`, `sin^2x-4!=0`

and hence we must have `cotx=0`.

Between `0` and `2pi`, `cotx=0` at `x=pi/2` and `x=(3pi)/2` and hence that is the solution,