Question

How do you find the sum of all solutions of `cosxcos(x+pi/3)cos(pi/3-x)=1/4`. if `x` is in `[0, 6pi]`?

Answer 1

Sum of all solutions is `30pi`.

Explanation:

`cosxcos(x+pi/3)cos(pi/3-x)=1/4` can be written as

`4cosxcos(x+pi/3)cos(pi/3-x)=1`

or `2cosx[2cos(x+pi/3)cos(pi/3-x)]=1`

or `2cosx[cos(x+pi/3+pi/3-x)+cos(x+pi/3-pi/3+x)]=1`

or `2cosx[cos((2pi)/3)+cos2x]=1`

or `2cosx[-1/2+2cos^2x-1]=1`

or `4cos^3x-3cosx=1`

or `cos3x=1=cos0`

i.e. `3x=2npi`, where `n` is an integer.

or `x=(2npi)/3`, where `n` is an integer.

and in `[0,6pi]` we will have

`x=[0,color(red)((2pi)/3,(4pi)/3),2pi,color(red)((8pi)/3,(10pi)/3),4pi,color(red)((14pi)/3,(16pi)/3),6pi]`

and their sum is `color(red)(2pi)+2pi+color(red)(6pi)+4pi+color(red)(10pi)+6pi=30pi`