How do you solve #Cos 2 theta = cos^2 theta - 1/2#?

1 Answer
Nghi N.
May 26, 2016

#pi/4, (3pi)/4, (5pi)/4, (7pi)/4#

Explanation:

Replace in the equation cos 2x by #(2cos^2 x - 1)# -->
#2cos^2 x - 1 = cos^2 x - 1/2#
#cos^2 x = 1 - 1/2 = 1/2#
#cos x = +- 1/sqrt2 = +- sqrt2/2#
Trig table and unit circle give -->
a. #cos x = sqrt2/2# --> #x = +- pi/4#
b. #cos x = -sqrt2/2# --> #x = +- (3pi)/4#
Answers for #(0, 2pi)# -->
#pi/4, (3pi)/4, (5pi)/4, (7pi)/4#

Note. Arc #-pi/4# is co-terminal to arc #(7pi)/4#
Arc #(-3pi)/4# is co-terminal to arc #(5pi)/4#

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