The correct option is C π3(6k+1), k∈Z
cos3x+sin(2x−7π6)=−2
⇒1+cos3x+1+sin(2x−7π6)=0
⇒(1+cos3x)+1−cos(2x−2π3)=0
⇒2cos23x2+2sin2(x−π3)=0
⇒cos3x2=0 and sin(x−π3)=0
⇒3x2=π2,3π2,⋯
and x−π3=0,π,2π,⋯
Therefore, the general solution is x=2kπ+π3=π3(6k+1), k∈Z