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

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

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Answer 1 · 2016-08-11T05:03:23

#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#

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

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