Take a look at the following function:
$$
x(t) = \sin^3(2t)
$$
In order to show the periodicity of the signal, we need to prove the following equality
$$
x(t) = x(t+T)
$$
We first use some trigonometric identities:
$$
x(t) = \sin^3(2t) = \frac{3}{4} \sin(2t)-\frac{1}{4}\sin(6t) \tag{1}
$$
What I've done is as follows:
$$
\begin{align}
x(t+T) &= \sin^3(2(t+T)) \\
&=\frac{3}{4}\sin(2(t+T)) – \frac{1}{4}\sin(6(t+T)) \\
&=\frac{3}{4}\sin(2(t+T)) – \frac{1}{4}\sin(6(t+T)) \\
&=\frac{3}{4}\sin(2t+2T) – \frac{1}{4}\sin(6t+6T) \\
&= \frac{3}{4}\Big[(\sin(2t)\cos(2T))+(\cos(2t)\sin(2T))\Big] – \\
&\frac{1}{4}\Big[(\sin(6t)\cos(6T))+(\cos(6t)\sin(6T))\Big] \tag{2}
\end{align}
$$
In order for the condition to hold, we choose $2T=2\pi \implies T=\pi$. Substituting $T$ in (2), we get
$$
\begin{align}
x(t+T) &=
\frac{3}{4}\Big[\sin(2t)\Big] – \frac{1}{4}\Big[\sin(6t)\Big] \\
&= x(t)
\end{align}
$$
So, the fundamental period is $T=\pi$ sec. The author states that the fundamental period is $T=1/\pi$ sec. Is it correct?
Best Answer
$T=1/\pi$ is definitely not the period of $s(t)$ as, for example,
$$x(2\cdot 0) = \sin^3(0) = 0 \neq x(2(0 + 1/\pi)) = \sin^3(2/\pi) \approx 0.2101$$
As noted in the comments, it's possible it was a typo or misprint or something since the period of $x(t)$ is indeed $\pi$, but we have the relation that frequency $f = 1/T$ so that might have been what is being referred to (as far as intention goes).
Whether that's the intention, I wouldn't know without having the book in front of me. You might have to just figure that bit out for yourself.