How do you solve #(sinx+1)-2cosx=0#?

1 Answer
Jun 22, 2016

#143^@30 and 269^@82#

Explanation:

Rewrite the equation:
sin x - 2cos x = - 1
Call t the arc that #tan t = sin t/(cos t) = -2# --> #t = 116^@56#
#sin x - (sin t)/(cos t)cos x = -1#
#sin x.cos t - sin t.cos x = -cos t = 0.45#
#sin (x - t) = sin (x - 116.56) = 0.45#.
There are 2 solution arcs:
a. #x - 116.56 = 26^@74# --> #x = 26.74 + 116.57 = 143^@30#
b. x - 116.56 = 180 - 26.74 = 153.26 -->
#x = 153.26 + 116.56 = 269^@82#

Check by calculator:
x = 143.30 --> sin x = 0.60 --> 2cos x = 1.60
0.60 - 1.60 = -1. OK
x = 269.82 --> sin x = -1 --> 2cos x = 0
-1 - 0 = -1 . OK