If for real values of $x, \cos \theta=x+\dfrac{1}{x}$, then

Given the equation $\cos \theta = x + \dfrac{1}{x}$ for real values of $x$.

For any real $x \neq 0$, the expression $x + \dfrac{1}{x}$ satisfies the inequality $|x + \dfrac{1}{x}| \ge 2$ by the Arithmetic Mean-Geometric Mean inequality (AM-GM) for positive $x$, and $|x + \dfrac{1}{x}| \le -2$ for negative $x$.

Specifically, if $x > 0$, then $x + \dfrac{1}{x} \ge 2$. If $x < 0$, let $x = -y$ where $y > 0$, then $x + \dfrac{1}{x} = -(y + \dfrac{1}{y}) \le -2$.

Thus, the range of the function $f(x) = x + \dfrac{1}{x}$ is $(-\infty, -2] \cup [2, \infty)$.

The range of $\cos \theta$ is $[-1, 1]$.

Since the range of $x + \dfrac{1}{x}$ and the range of $\cos \theta$ have no common values except possibly at the boundaries, we check if $\cos \theta$ can equal $2$ or $-2$. Since $\cos \theta$ cannot exceed $1$ or be less than $-1$, there is no real value of $x$ for which $\cos \theta = x + \dfrac{1}{x}$ holds true.

Answer: No value of $\theta$ is possible