How do you find all values of x in the interval [0, 2pi] in the equation # (cos x+1) (2cos^2 x-3 cos x-2) =0#?

1 Answer
Nghi N.
Mar 15, 2016

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

Explanation:

Solve the 2 factors:
a. (cos x + 1) = 0 --> cos x = -1 --> #x = pi#
b. #2cos^2 x - 3cos x - 2 = 0#.
Use the improved quadratic formula to solve for cos x.
#D = d^2 = b^2 - 4ac = 9 + 16 = 25# --> #d = +- 5#
#cos x = -b/(2a) +- d/(2a) = 3/4 +- 5/4#.
#cos x = 8/4 = 2# (rejected since > 1), and #cos x = - 2/4 = -1/2#
#cos x = - 1/2# --># x = +- (2pi)/3# -->
#x = (2pi)/3#, and #x = (4pi)/3# (co-terminal to #-(2pi)/3#)
Answers for #(0, 2pi):#
#(2pi)/3, pi, (4pi)/3#

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