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

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

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 #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.

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