Since the equation is transcendental, to solve it for $x$ would require a numerical method.
However, you could use the beautiful approximation
$$\cos(x) \sim\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad \text{for} \qquad -\frac \pi 2 \leq x\leq\frac \pi 2$$ derived from the sine approximation proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (have a look here).
This is just a quadartic equation in $x^2$ and the approximate solution is then
$$x=\sqrt{\frac{1}{6} \left(\sqrt{16+9 \pi ^2 \left(4+\pi
^2\right)-(4+3 \pi ^2)}\right)}\approx 0.535110$$ while the exact solution, obtained using Newton method, is $0.535428$ (then corresponding to a relative error of $0.06$%).
Another way would based on inspection : graphing, you see that the solution is close to $\frac \pi 6$. So, using Taylor expansion
$$y=3x^2-\cos(x)=\frac{1}{12} \left(\pi ^2-6 \sqrt{3}\right)+\left(\frac{1}{2}+\pi \right)
\left(x-\frac{\pi }{6}\right)+\left(3+\frac{\sqrt{3}}{4}\right)
\left(x-\frac{\pi }{6}\right)^2-\frac{1}{12} \left(x-\frac{\pi
}{6}\right)^3+O\left(\left(x-\frac{\pi }{6}\right)^4\right)$$ and, using series reversion
$$x=\frac{\pi }{6}+t-\frac{\left(12+\sqrt{3}\right) t^2}{2 (1+2 \pi
)}+\frac{\left(221+36 \sqrt{3}+\pi \right) t^3}{3 (1+2 \pi
)^2}+O\left(t^{4}\right)$$ where $t=\frac{12 y-\pi ^2+6 \sqrt{3}}{6(1+2 \pi) }$
Making $y=0$ and computing leads to $x=0.535428330$ while the exact solution is $0.535428244$
Edit
Even simpler could be obtained using, instead of the beautiful approximation proposed by Mahabhaskariya of Bhaskara I, the simplest Padé approximant
$$\cos(x)=\frac{12-5 x^2}{12+x^2}$$ which, for sure, is less accurate. Again, a quadratic equation in $x^2$ and the solution of it is
$$x=\sqrt{\frac{5 \sqrt{73}-41}{6}}\approx 0.535415528$$ Using the next Padé approximant
$$\cos(x)=\frac{3 x^4-56 x^2+120}{4 x^2+120}$$ would give
$$x=\frac{1}{3} \sqrt{2 \left(\sqrt{11086}-104\right)}\approx 0.535428283$$
