Question

How do you find all solutions for `sinx - tanx = 0` where x is between 0 and 2(pi)?

Answer 1

`f(x) = sin x - sin x/cos x = sin x*(1 - 1/cos x) = sin x*(cos x - 1)/cos x.`
Solve `sin x = 0` and `(cos x - 1) = 0`
Reminder. `f(x)` undefined when `cos x = 0 --> x = pi/2 and (3pi)/2.`
`sin x = 0 --> x = 0; x = pi`; and `x = 2pi`
`cos x - 1 = 0 --> cos x = 1 --> x = 0; and x = 2pi`
Answers within period `(0, 2pi): 0, pi, 2pi`.
Check:
`x = pi --> f(x) = 0 - 0 = 0` (correct)