$\begin{aligned}|x(t)|^{2} & =4 \cos ^{2}(2 n t+\pi / 3) \cdot[u(t+1)-u(t-2)] \\ & =2(1+\cos (4 \pi t+2 \pi / 3)[u(t+1)-u(t-2)]\end{aligned}$

Step 1 ：Given the equation $|x(t)|^{2} =4 \cos ^{2}(2 n t+\pi / 3) \cdot[u(t+1)-u(t-2)] =2(1+\cos (4 \pi t+2 \pi / 3)[u(t+1)-u(t-2)]$

Step 2 ：We can see that the equation is defined for $t \in [-1, 2]$ due to the unit step functions $u(t+1)$ and $u(t-2)$.

Step 3 ：We can simplify the equation by using the identity $\cos^2(x) = \frac{1 + \cos(2x)}{2}$, which gives us $|x(t)|^{2} = 2(1+\cos (4 \pi t+2 \pi / 3))$

Step 4 ：Taking the square root of both sides, we get $|x(t)| = \sqrt{2(1+\cos (4 \pi t+2 \pi / 3))}$

Step 5 ：Since $|x(t)|$ is always non-negative, we can say that $x(t) = \sqrt{2(1+\cos (4 \pi t+2 \pi / 3))}$ for $t \in [-1, 2]$

Step 6 ：This is the solution to the given equation.