Question

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

Answer 1

#(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]`

Explanation:

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`

Answer 2

`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`