My attempts:
First of all, if we draw graphs of left and right sides of the equation, will see that $x(t) \rightarrow 0$ as $t \rightarrow \infty$, because segment $[0, \frac{\pi}{2t}]$ shrinks with growth of $t$. It's obvious.
Of course, it is too little and we need at least to find out an infinitesimal order of $x(t)$.
I thought about expansion $\cos(tx)$ into a Taylor series: $$\frac12 + x^2 = 1 - \frac{(tx)^2}{2!} + \frac{(tx)^4}{4!} - \dots$$ but it didn't help much.
And some special cases:
If $t = 0$, then $x = \frac{1}{\sqrt2}$.
If $t = 1$, then $\frac12 + x^2 = 1 - \frac{x^2}{2!} + O(x^4)$ and $x \approx \frac{1}{\sqrt3}$.
If $t = 2$, then $\frac12 + x^2 = 1 - \frac{4x^2}{2!} + O(x^4)$ and $x \approx \frac{1}{\sqrt6}$.
But with growth of $t$ we can't continue with this idea.
I guess, we need to check hypotheses about the order of $x(t)$ and look if there is any contradiction with Taylor series, but I got stuck there.
Denote $u(t)=tx(t)$. Then $u$ is the least positive root of $$ F_t(u)=\frac{u^2}{t^2}+\frac12-\cos u. $$ We can see that $F_t(0)=-\frac12$ and $F_t(\frac{\pi}3)>0$. Since $F_t'(u)=\frac{2u}{t^2}+\sin u>0$ on $[0,\frac{\pi}3]$, it follows by intermediate value theorem and Rolle's theorem, that there is a unique root $u(t)$ of $F_t$ on $(0,\frac{\pi}3)$. Also, $\partial_tF_t(u)<0$ implies that $\frac{d}{dt}u(t)=-\frac{\partial_t F_t(u)}{F_t'(u)}>0$, i.e. $u(t)$ is increasing by implicit function theorem. Thus $u^*=\lim_{t\to\infty}u(t)\in [0,\frac{\pi}3]$ exists by monotonicity and should satisfy $$ \frac12=\cos u^*, $$that is, $u^*=\lim_{t\to\infty}tx(t)=\frac{\pi}3$.