How do you solve #2cos^2x+cosx -1=0# for #0<=x<=2pi#?

1 Answer
Nghi N.
Jun 21, 2018

#pi/3; pi; (5pi)/3#

Explanation:

#2cos^2 x + cos x - 1 = 0#
Solve this quadratic equation for cos x.
Since a - b + c = 0, use shortcut. The 2 real roots are:
cos x = -1, and #cos x = -c/a = 1/2#
a. cos x = -1
Unit circle gives:
#x = pi#
b. #cos x = 1/2#
Trig table and unit circle give -->
#x = +- pi/3 #
Note.
#x = - pi/3# is co-terminal to #x = (5pi)/3#
Answers for #(0, 2pi)#
#pi/3; pi; (5pi)/3#

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