Question

How do you solve `1-sinx=sqrt3cosx`?

Answer 1

`2kpi`
`pi/3 + 2kpi`

Explanation:

`1 - sin x = sqrt3cos x`
`sin x + sqrt3cos x = 1`
Call t the arc whose `tan t = sqrt3` -->`t = pi/3`
We get:
`sin x + (sin t)/(cos t)cos x = 1`
`sin x.cos (pi/3) + sin (pi/3).cos x = cos t = 1/2`
`(sin x + pi/3) = 1/2`
Trig table and unit circle give 2 solution arcs:
a. `x + pi/3 = pi/3` --> x = 0
b. `x + pi/3 = 2pi/3` --> `x = (2pi)/3 - pi/3 = pi/3`
General answers:
`x = 2kpi`
`x = pi/3 + 2kpi`