Question

How do I solve 1-sinx = 3cosx for [0,2pi)?

`1-sinx = 3cosx` for `[0,2pi)`

Answer 1

`x = 89^@77`
`x = 307^@09`

Explanation:

1 - sin x = 3cos x
sin x + 3cos x = 1 (1)
Call `tan t = 3 = sin t/(cos t) = tan 71^@ 57` --> cos t = 0.32
Equation (1) becomes:
sin x.cos t + sin t.cos x = cos t = 0.32
sin (x + t) = sin (x + 71.57) = 0.32
Calculator and unit circle give 2 solutions -->

a. `x + 71.57 = 18^@66` -->
`x = 18.66 - 71.57 = - 52^@91`, or,
`x = 307^@09` (co-terminal to - 52.91)
b. `x + 71.57 = 180 - 18.66 = 161^@34` -->
`x = 161.34 - 71.57 = 89^@77`
Check by calculator.
x = 307.09 --> sin x = - 0.8 --> 1 - sin x = 1 + 0.8 = 1.8
cos x = 0.6 --> 3cos x = 1.8. Proved.