If sum of all the solutions of the equation 8cosx.(cos(π/6+x).cos(π/6-x))=1 in [0,π] is kπ, then k is equal to

8cosx[cos(π/6+x)cos(π/6-x)]=1
using 2cosAcosB=cos(A+B)+cos(A-B), we get
8cosx[cos(π/3)+cos(2x)/2]=1
∴4cosx[1/2+cos(2x)/2]=1
∴4cosx[cos(2x)/2]=1
∴4cosxcos(2x)=1
using 2cosAcosB=cos(A+B)+cos(A-B), we get
2(cos3x+cosx)=1
∴cos3x=1/2
∴3x=2nπ±π/3
∴x=2nπ/3±π/9
Solutions in [0,π] are π/9, 2π/3-π/9, 2π/3+π/9
Hence, their sum=π/9+5π/9+7π/9=13π/9
∴k=13/9
This is the required solution.