How do you solve # sin (x/2) + cosx -1= 0# over the interval 0 to 2pi?

1 Answer
Nghi N.
Jan 31, 2016

#0, 2pi, pi/3; 5pi/3 #-->interval #(0, 2pi)#

Explanation:

Use trig identity: #cos 2a = 1 - 2sin^2 a#
#cos x = 1 - 2sin^2 (x/2)#.
Call #sin (x/2) = t#, we get:
#t + (1 - 2t^2) - 1 = t(1 - 2t) = 0#
2 solutions:
a. #t = sin (x/2) = 0 #-->
#x/2 = 0# --> #x = 0#, and
#x/2 = pi# --> #x = 2pi#
b. 1 - 2t = 0 --> #t = sin x = 1/2#
#t = sin x/2 = 1/2# --> 2 answers:
#x/2 = pi/6# --> #x = pi/3#
#x/2 = (5pi)/6 --> x = (5pi)/3#
Check
x = pi/3 --> x/2 = pi/6 --> sin x/2 = 1/2 --> cos pi/3 = 1/2 -->
1/2 + 1/2 - 1 = 0. Correct
x = (5pi)/3 --> x/2 = (5pi)/6 --> sin (5pi)/6 = 1/2 --> cos (5pi)/3 = cos (pi/3) = 1/2 --> 1/2 + 1/2 - 1 = 0. OK

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