f(t)=acosb(t-c)+d

maximum value is at t=3 but for maximum value cosb(t-c) should be maximum that is 1 therefor

29=a+d............................................ ...1

minimum value at t=9 but for minimum value cosb(t-c) should be minimum that is -1 therefor

15=-a+d............................................... 2

solving equation 1 and 2 we get

a=7

b=22

now

f(t)=7cosb(t-c)+22

since max value was 29 at t=3 therefor

29=7cosb(3-c)+22 or

7=7cosb(3-c) or

cosb(3-c)=1 or

b(3-c)=0.................................3 (cosx=1 then x=0)

similarly we have

15=7cosb(9-c)+22 or

cosb(9-c)=-1 or

b(9-c)=pi..................................4 (cosx=-1 then x=pi)

solving equation 3 and 4 simultaneously we get

b=pi/6 hence proved