The function $f(x)=3\sin^2\pi x+2x-\lfloor2x\rfloor$ is defined in real numbers. What is the smallest period of the function?
$1)1\qquad\qquad2)2\qquad\qquad3)\pi\qquad\qquad4)2\pi$
Period of $2x-\lfloor 2x\rfloor$ is $\frac12$ and I think period of $\sin^2\pi x=\cfrac{1-\cos(2\pi x)}{2}$ is $\dfrac{2\pi}{2\pi}=1$.
Hence period of $f(x)$ is $1$. Am I right?