Introduction
In this problem, we are given the functions sin(2πx/a) and 3cos(2πx/b), and we are asked to find their period when a=12 and b=9.

Finding the Period of sin(2πx/a)
The period of the function sin(2πx/a) is given by the formula 2πa/ω, where ω is the angular frequency of the function. Since sin(2πx/a) has an angular frequency of 2π/a, its period is given by:

T1 = 2πa/(2π/a) = 2a

Therefore, the period of sin(2πx/12) is 2×12 = 24.

Finding the Period of 3cos(2πx/b)
Similarly, the period of the function 3cos(2πx/b) is given by the formula 2πb/ω, where ω is the angular frequency of the function. Since cos(2πx/b) has an angular frequency of 2π/b, its period is given by:

T2 = 2πb/(2π/b) = 2b

Therefore, the period of 3cos(2πx/9) is 2×9 = 18.

Finding the Period of the Product Function
To find the period of the product function sin(2πx/12)×3cos(2πx/9), we need to find the least common multiple (LCM) of the periods of the two functions. The LCM of 24 and 18 is 72, so the period of the product function is 72.

Conclusion
Therefore, the period of the function sin(2πx/12)×3cos(2πx/9) is 72.