The correct option is C [−4,−3]
cos(4y−3x−2)−cos(4y+3x+2)=2+2ln(k4−255)cos(4y−3x−2)+cos(4y+3x+2)=2k+8
Assuming 4y=A,3x+2=B, we get
cos(A−B)−cos(A+B)=2(1+ln(k4−255))⇒2sinAsinB=2(1+ln(k4−255))⇒1+ln(k4−255)=sinAsinB⇒1+ln(k4−255)≤1⇒ln(k4−255)≤0⇒k4−255≤1⇒k4≤256⇒k∈[−4,4]⋯(1)
Also,
cos(A−B)+cos(A+B)=2k+8⇒2cosAcosB=2(k+4)⇒k+4=cosAcosB⇒k+4≤1⇒k≤−3⋯(2)
From equation (1) and (2), we get
k∈[−4,−3]