How do you solve: #Cos (x/2) - cosx=1#?

1 Answer
Nghi N.
Mar 19, 2016

#pi, (2pi)/3, (4pi/3)#

Explanation:

Apply the trig identity: #cos 2a = 2cos^2a - 1#
#cos x = 2cos^2 (x/2) - 1#. We get:
#cos (x/2) - 2cos^2 (x/2) + 1 = 1#
#cos (x/2)( 1 + 2cos (x/2)) = 0#
a. #cos x/2 = 0# --> 2 solutions:
#x/2 = pi/2# --> #x = pi#
#x/2 = 3pi/2# --> #x = 3pi# or #x = pi#.
b. #2cos (x/2) = - 1# --> #cos (x/2) = - 1/2# -->
Trig table --> 2 solutions:
#x/2 = (2pi)/3# --> #x = (4pi)/3#
#x/2 = (4pi)/3# --> #x = (8pi)/3#, or #x = (2pi)/3#
Answers for #(0, 2pi):#
#pi, (2pi)/3, (4pi)/3#

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