We know that,
color(violet)((I)cosC+cosD=2cos((C+D)/2)cos((C-D)/2)(I)cosC+cosD=2cos(C+D2)cos(C−D2)
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