Question

How do you solve `sinx-tanx=0` from [0,2pi]?

Answer 1

I found: `x=0,pi,2pi`

Explanation:

Write it as:
`sin(x)-sin(x)/cos(x)=0` collect `sin(x)`
`sin(x)[1-1/cos(x)]=0`
So you get:
`sin(x)=0` for `x=0, pi, 2pi`
`1-1/cos(x)=0` or `cos(x)=1` for `x=0, 2pi`