Without Calculator find $\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$ 

Without Calculator find $$\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$$
Where $\left \lfloor x \right \rfloor $ represents floor function.


My Try:
Let $x=2\cos(50^{\circ})+\sqrt{3}$. We have
$$\begin{aligned}
& \cos \left(50^{\circ}\right)<\cos \left(45^{\circ}\right) \\
\Rightarrow \quad & 2 \cos \left(50^{\circ}\right)<\sqrt{2} \\
\Rightarrow \quad & x<\sqrt{3}+\sqrt{2}<3.14
\end{aligned}$$
Now I am struggling hard to prove that $x>3$

Without derivatives:
Remark: I used this trick in my answer before.
We want to prove that $2\cos \frac{50\pi}{180}  + \sqrt 3 > 3$.
Let $u = \cos \frac{50\pi}{180}$ and $v = \frac{3 - \sqrt 3}{2}$. We need to prove that $u > v$.
We have $u \ge \cos \frac{60\pi}{180} = 1/2$ and $v > \frac{3 - 2}{2} = 1/2$.
Thus, $4u^2 + 4uv + 4v^2 > 3$.
Thus, it suffices to prove that
$$(u - v)[4u^2 + 4uv + 4v^2 - 3] > 0$$
or
$$4u^3 - 3u > 4v^3 - 3v$$
or (using triple angle formula $\cos 3x = 4\cos^3 x - 3\cos x$)
$$\cos \left(3 \times \frac{50\pi}{180}\right) > \frac{45}{2} - \frac{27}{2}\sqrt 3$$
or
$$13\sqrt 3 > \frac{45}{2}$$
which is true.
We are done.
