Lets find the intersection of the curves in the first quadrant:
3costheta=1+costheta => 2costheta=1 => costheta=1/2 => theta=pi/3
The region is symmetric so we can find the area of the half of it:
A= 2 (int_0^(pi/3) d theta int_0^(1+costheta) rdr + int_(pi/3)^(pi/2) d theta int_0^(3costheta) rdr)
A_1= 1/2int_0^(pi/3) d theta r^2 |_0^(1+costheta) = 1/2int_0^(pi/3) d theta (1+2costheta+cos^2theta)
A_1= 1/2 int_0^(pi/3) d theta (1+2costheta+(1+cos2theta)/2)
A_1= 1/2 ((3theta)/2+2sintheta+1/4sin2theta)|_0^(pi/3) =pi/4+(9sqrt3)/16
A_2= 1/2int_(pi/3)^(pi/2) d theta r^2 |_0^(3costheta)=
1/2int_(pi/3)^(pi/2) d theta (9cos^2theta) = 9/4 int_(pi/3)^(pi/2) d theta (1+cos2theta) =
= 9/4 (theta +1/2sin2theta) |_(pi/3)^(pi/2) = (3pi)/8-(9sqrt3)/16
A=2(pi/4+(9sqrt3)/16 + (3pi)/8-(9sqrt3)/16) = 2(5pi)/8 = (5pi)/4