Answer
The equation $~~2x+cos~x = 0~~$ has exactly one real root.
Work Step by Step
$2x+cos~x = 0$
Let $f(x) = 2x+cos~x$
The function $f(x) = 2x+cos~x$ is continuous and differentiable for all $x$.
When $x=-1$, then $(2x+cos~x) \lt 0$
When $x=1$, then $(2x+cos~x) \gt 0$
Therefore, by the Intermediate Value Theorem, there is a number $c$ in the interval $(-1,1)$ such that $[2(c)+cos~(c)] = 0$.
Thus the equation has at least one real root.
Let's assume that the equation has at least two real roots $a$ and $b$. Then $f(a) = f(b) = 0$. According to Rolle's Theorem, there is a number $k$ such that $f'(k) = 0$.
We can try to find $k$:
$f'(x) = 2-sin~x = 0$
$sin~x = 2$
However, there is no number $x$ such that $sin~x = 2$. Therefore, the assumption that the equation has at least two real roots is incorrect.
Therefore, the equation has exactly one real root.