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

2 Answers
May 12, 2018

(i)x=pi/7,(3pi)/7,(5pi)/7,(7pi)/7,(9pi)/7,(11pi)/7,(13pi)/7.tox in [0,2pi](i)x=π7,3π7,5π7,7π7,9π7,11π7,13π7.x[0,2π]
(ii)x=pi/3,(3pi)/3,(5pi)/3.to x in [0,2pi](ii)x=π3,3π3,5π3.x[0,2π]

Explanation:

We know that,

color(violet)((I)cosC+cosD=2cos((C+D)/2)cos((C-D)/2)(I)cosC+cosD=2cos(C+D2)cos(CD2)

Here,

cos5x=-cos2xcos5x=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

May 13, 2018

pi/3 ; pi ; (5pi)/3
pi/7 ; (3pi)/7 ; (5pi)/7 ; pi; (9pi)/7 ; (11pi)/7 ; (13pi)/7

Explanation:

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