How do you solve #2cos^2x - cosx-1=0# in the interval 0<=x<2pi#?

1 Answer
Nghi N.
Jul 26, 2016

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

Explanation:

#f(x) = 2cos^2 x - cos x - 1 = 0#
Solve this quadratic equation for cos x.
Since a + b + c = 0, use shortcut.
One real root is cos x = 1 and the other is #cos x = c/a = - 1/2#.
Trig table and unit circle -->
a. cos x = 1 --> arc x = 0, and arc #x = 2pi#
b. #cos x = - 1/2# --> arc #x = +- (2pi)/3#
The arc #(-2pi)/3# is co-terminal to the arc: #(4pi)/3#
Answers for #(0, 2pi):#
#0, (2pi)/3, (4pi)/3, and 2pi#

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