Solve the equation #cosx(1+2cosx)=0#?

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Answer 1 · 2017-05-22T04:55:10

#x = kpi+pi/2#
#x = +- (2pi)/3 + 2kpi#, where #k# is an integer.

#cos x(1 + 2cos x) = 0#

Either #cos x# or #(1 + 2cos x)# must be zero.

Use unit circle:

a. #cos x = 0 --> x = pi/2# or #x = (3pi)/2#

General answer: #x = kpi+pi/2#, where #k# is an integer

b. #1 + 2cos x = 0 -> cos x = - 1/2#

Use trig table and unit circle:

#x = +- (2pi)/3 + 2kpi#