How do you solve #cos(x/2) - sin(x) = 0#?

1 Answer
Nghi N.
Apr 30, 2016

#pi + 2kpi#
#2kpi#
(5pi)/3 + 2kpi

Explanation:

Use trig identity: #sin x = 2sin (x/2).cos (x/2)#. We get:
#cos (x/2)- sin (x/2).cos (x/2) = 0#
#cos (x/2)(1 - 2sin x) = 0#
a. #cos (x/2) = 0 # -->arc #x/2 = pi/2# and arc #x = (3pi)/2# -->
#x = pi# and #x = 3pi.#
General answer -->#x = pi + 2kpi#
b.#sin (x/2) = 1/2#.
Trig table and unit circle give:
#x/2 = pi/6# and #x/2 = (5pi)/6# -->
#x = (2pi)/6# and #x = (5pi)/3#
General answers:
#x = pi + 2kpi#
#x = 2kpi#
#x = (5pi)/3 + 2kpi#

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