How do you solve #cos (theta/2) - cos theta = 1# over the interval 0 to 2pi?

1 Answer
Nghi N.
May 21, 2016

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

Explanation:

Use the trig identity to transform cos t to cos (t/2) -->
#cos t = 2cos^2 (t/2) - 1.#
The equation becomes:
#cos (t/2) - 2cos^2 (t/2) + 1 = 1#
#cos (t/2) - 2cos^2 (t/2) = 0#
Put #cos (t/2)# into common factor -->
#cos (t/2)(1 - 2cos (t/2)) = 0#
Trig table and unit circle -->
a. #cos (t/2) = 0# --> There are 2 solution arcs:
#(t/2) = pi/2# --> #t = (2pi)/2 = pi#
#t/2 = (3pi)/2# --># t = (6pi)/2 = 3pi = pi#
b. #cos (t/2) = 1/2# --> #t/2 = +- pi/3 #-->
#t/2 = pi/3# --> #t = (2pi)/3#
and #t/2 = - pi/3# or #t/2 = (5pi)/3# (co-terminal of #-(pi/3))#-->
#t = (10pi)/3 = ((4pi)/3 + 2pi) = (4pi)/3#

Answers for #(0, 2pi):#
#pi, (2pi)/3, (4pi)/3 #
Checking these answers by calculator is advised.

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