Question

How do you solve `tan(x/2) - sinx = 0` over the interval 0 to 2pi?

Answer 1

`0, pi/4, pi/2, (3pi)/4, (5pi)/4, (3pi)/2, (7pi)/4 and 2pi`

Explanation:

Call `x/2 = t` . Apply the identity: `sin 2a = 2sin a.cos a`
`(sin t)/(cos t) - 2sin t.cos t = 0`.
`(sin t)/(cos t) - (2sin t.cos t) = 0`
`sin t(1 - 2cos^2 t) = 0`
a. `sin t = sin (x/2) = 0` --> `x/2 = 0, x/2 = pi and x/2 = 2pi` -->
`x = 0 and x = 2pi`
b. `1 - 2cos^2 t = 0` --> `cos^2 t = 1/2` -->
`cos t = cos (x/2) = +-sqrt2/2`
`cos t = cos (x/2) = sqrt2/2 --> x/2 = +- pi/4 --> x = +-pi/2`
`cos t = cos (x/2) = -sqrt2/2` --> `x/2 = +-(3pi)/4` --> `x = +-3pi/2`
Answers for `(0, 2pi)`
`0, pi/4, pi/2, (3pi)/4, (5pi)/4, (3pi)/2, (7pi)/4, 2pi.`
Note.
`- pi/4` --> `(7pi)/4` (co-terminal)
`-(3pi)/4` --> `(5pi)/4` (co-terminal)