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 $x>3$

50° is between 45° and 60° ( $50°=45°+(60°-45°)/3$ ), and in this range, function $cos$ is concave. So $\cos(50°) > \cos 45°+(\cos60°-\cos45°)/3$
$\cos 50° > \sqrt{2}/2 + (1/2- \sqrt{2}/2)/3$.
Approximate $\sqrt{2}$ with $1.414$ and $\sqrt{3}$ with $1.732$, and you are in.
