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

1−sinx=3cosx for [0,2π)

1 Answer
Nghi N
Mar 18, 2018

x=89∘77
x=307∘09

Explanation:

1 - sin x = 3cos x
sin x + 3cos x = 1 (1)
Call tant=3=sintcost=tan71∘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.

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