How do you solve #cos2x= cosx# from 0 to 2pi?

1 Answer
Nghi N.
Apr 11, 2016

# 0, (2pi)/3, (4pi)/3, 2pi#

Explanation:

Use the identity: #cos 2x = 2cos^2 x - 1#. The given equation
transforms to:
#2cos^2 x - cos x - 1 = 0#.
Solve this quadratic equation for cos x.
Since a + b + c = 0, use shortcut. There are 2real roots:
cos x = 1 and #cos x = c/a = - 1/2#.
a. cos x = 1 --> x = 0 or #x = 2pi#
b. #cos x = - 1/2# ---> #x = +- (2pi)/3#
The co-terminal to arc #- (2pi)/3# --> arc #(4pi)/3#

Answers for #(0, 2pi)#:
#0, (2pi)/3, (4pi)/3, 2pi#

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