As for every $t_0\in\mathbb{R}$ and $k\in\mathbb{Z}$

$$\begin{eqnarray} &x(t_0+4k\pi) &=\cos(t_0+4k\pi)+\sin(t_0/2+2k\pi)\\ & &=\cos(t_0)+\sin(t_0/2)\\ & &=x(t_0) \end{eqnarray}$$ you answer is correct: $x(t)$ is periodic.

To add a contrarian answer: If your time index, $t$, is an integer, then your signal is not periodic.

The definition of periodic is: $x[t]$, $t\in\mathbb{Z}$ is periodic with period $P\in \mathbb{Z}$ iff

$$x[t] = x[t+P]$$

So we need

$$\cos(t) = \cos(t+P)$$ so for periodicity we require $$P = 2\pi k$$ with $k\in \mathbb{Z}$.

Since $\pi$ is irrational, that cannot be the case.

Hence the first component of your signal cannot be periodic, so the entire signal cannot be periodic.

The double-angle formulae for trigonometric identities tell you that $\cos \left(\frac{2t}{t}\right) = 1 - 2\sin^2(\frac{t}{2})$.

You thus ave $x(t) =1 +\sin(\frac{t}{2}) - 2\sin^2(\frac{t}{2})$. Hence, your signal is composed of functions (as adds and multiplies) that all admit $4\pi$ as a period (yes, the constant function $x\to 1$ is $4\pi$ periodic as well).

Thus your function looks very periodic.