How do i solve 1=2cos(3x-5pi) for [0,2)?

#1=2cos(3x-5pi)# for #[0,2pi)? #

1 Answer
Nghi N
Feb 14, 2018

#x = +- (2pi)/9#, or
#x = +- 40^@#

Explanation:

#cos (3x - 5pi) = 1/2#
#cos (3x - 5pi) = cos (3x - pi) = - cos 3x# -->
#cos 3x = - 1/2#
Trig table and unit circle give 2 solutions:
#3x = +- (2pi)/3#
#x = +- (2pi)/9#
Check by calculator:
#x = (2pi)/9 = 40^@# --> #2cos (3x - pi) = 2cos (120 - 180) =#
#= 2cos (-60) = 2cos 60 = 2(1/2) = 1# Proved.

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