Question #af5d8

1 Answer
Nghi N
May 2, 2017

a. cos 2x = 0
Unit circle gives:
#2x = pi/2 + 2kpi# and #2x = (3pi)/2 + 2kpi#

#2x = pi/2 + 2kpi# --> #x = pi/4 + kpi#
#2x = (3pi)/2 + 2kpi# --> #x = (3pi)/4 + kpi#
Answers for #(0, 2pi)#:
#pi/4; (3pi)/4; (5pi)/4; (7pi)/4#

b. #cos x = cos 2x = 2cos ^2 x - 1#
Solve this quadratic equation for cos x:
#2cos^2 x - cos x - 1 = 0#
Since a + b + c = 0, use shortcut.
The 2 real roots are: cos x = 1 and cos x = c/a = - 1/2
Trig table and unit circle give:
1/ cos x = 1 --> x = 0 and x = 2kpi
2/ #cos x = - 1/2# --> #x = +- (2pi)/3 + 2kpi#
Answers for #(0, 2pi)#:
#2pi; +- (2pi)/3#
c. sin 3x = 0
Unit circle gives as solutions:
1/ #3x = 0 + 2kpi# --> #x = (2kpi)/3#
2/ #3x = pi + 2kpi #--> #x = ((2k + 1)pi)/3#
3/ 3x = 2pi + 2kpi --> x = (2kpi)/3
Answers for #(0, 2pi)#:
#(2pi)/3; (4pi)/3; 2pi; pi/3; pi#

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