I think it is 3. You therefore know that if your cosine portion and your sine portion started off at a certain value each, both will be back at that certain value after t changes by 3.
No! If t changes by 3 then you have
f(t+3)=\cos3\,\pi\,(t+3) + \frac{1}{2}\,\sin4\,\pi\, (t+3)=\cos(3\,\pi\,t+9\,\pi)+\frac{1}{2}\,\sin(4\,\pi\,t+12\,\pi)=-\cos(3\,\pi\,t)+\frac{1}{2}\,\sin(4\,\pi\,t+12\,\pi)\neq f(t)
so the period
is not 3.
In order to find the period of the above function you have to work like this.
The period of the functions h(x)=\sin\alpha\,x,\,g(x)=\cos\alpha\,x is T=\frac{2\,\pi}{|\alpha|}. Of course every multiple of T is a period too, in the sense that the functions repeat their values after n\,T,\,n\in \mathbb{N^*}. For the function f the period of cosine is T_1=\frac{2}{3}\quad\text{or}\quad\tau_1=\,n\frac{2}{3} and for the sine T_2=\frac{1}{2}\quad\text{or}\quad\tau_2=m\frac{1}{2}. Now in order for the two functions to have the same period, their periods must be equal, thus
n\frac{2}{3}=m\frac{1}{2}\Rightarrow n=\frac{3}{4}\,m
Thus the first integer value of m which makes n also an integer is m=4\Rightarrow n=3 and the period of f is T=2.
