How do you solve $1 - sin x = \sqrt{3} cos x$ $1-sinx=sqrt3cosx$ ?

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How do you solve $1 - sin x = \sqrt{3} cos x$ $1-sinx=sqrt3cosx$ ?

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Answer 1 · 2016-08-11T05:03:23

$2 k π$ $2kpi$
$\frac{π}{3} + 2 k π$ $pi/3 + 2kpi$

Explanation:

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

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