How to find the exact solutions which lie in [0, 2π) for cos(5x) = −cos(2x)?

Question

Josiah Anderson · Accepted Answer

#pi/3 ; pi ; (5pi)/3#
#pi/7 ; (3pi)/7 ; (5pi)/7 ; pi; (9pi)/7 ; (11pi)/7 ; (13pi)/7# cos 5x = - cos 2x #cos 5x = cos (2x + pi)# Unit circle and property of cos x --> #5x = +- (2x + pi) + 2kpi# a. #5x = 2x + pi + 2kpi = 2x + (2k + 1)pi# #3x = (2k + 1)pi# #x = (2k +1)pi/3# k = 0 --> #x = pi/3# ; k = 1 --> #x = pi# ; k = 2 --> #x = (5pi)/3# b. #5x = - 2x - pi + 2kpi = - 2x + (2k - 1)pi# #7x = (2k - 1)pi# #x = (2k - 1)pi/7# k = 0 --> #x = - pi/7# or #x = (13pi)/7# (co- terminal) k = 1 --> #x = pi/7# ; k = 2 --> #x = (3pi)/7# ; k = 3 --> #x = (5pi)/7# ; k = 4 --> #x = pi# ; k = 5 --> #x = (9pi)/7# ; k = 6 --> #x = (11pi)/7#

Aaliyah Abdullah · Answer

#(i)x=pi/7,(3pi)/7,(5pi)/7,(7pi)/7,(9pi)/7,(11pi)/7,(13pi)/7.tox in [0,2pi]#
#(ii)x=pi/3,(3pi)/3,(5pi)/3.to x in [0,2pi]# We know that, #color(violet)((I)cosC+cosD=2cos((C+D)/2)cos((C-D)/2)# Here, #cos5x=-cos2x# #=>color(violet)(cos5x+cos2x=0...toApply(I)# #=>color(violet)(2cos((5x+2x)/2)cos((5x-2x)/2)=0# #=>cos((7x)/2)cos((3x)/2)=0# #=>color(blue)(cos((7x)/2)=0 or cos((3x)/2)=0# #color(blue)((i)cos((7x)/2)=0# #=>(7x)/2=pi/2,(3pi)/2,(5pi)/2,(7pi)/2,(9pi)/2,(11pi)/2,(13pi)/2, (15pi)/2,...# #=>7x=pi,3pi,5pi,7pi,9pi,11pi,13pi,15pi,...# #=>color(red)(x=pi/7,(3pi)/7,(5pi)/7,(7pi)/7,(9pi)/7,(11pi)/7,(13pi)/7.tox in [0,2pi]# #color(blue)((ii)cos((3x)/2)=0# #=>(3x)/2=pi/2,(3pi)/2,(5pi)/2,(7pi)/2,(9pi)/2,...# #=>3x=pi,3pi,5pi,7pi,9pi,...# #=>color(red)(x=pi/3,(3pi)/3,(5pi)/3.to x in [0,2pi]# Note: The general solution is: #=>(7x)/2=(2k+1)pi/2,kinZZ or(3x)/2=(2k+1)pi/2,kinZZ# #=>x=(2k+1)pi/7,kinZ or x=(2k+1)pi/3,kinZ#