Show that the equation has exactly one real root. 2x+\cos x=0

Show that the equation has exactly one real root.
$$\displaystyle{2}{x}+{\cos{{x}}}={0}$$

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FieniChoonin

First, remember the intermediate value theorem which states that given a function f(x)
continuous on [a, b] then there exists c such that : f(b) < f(c) < f(a) or f(a) < f(c) < f(b)
$$\displaystyle{2}{x}+{\cos{{x}}}={0}$$
Let $$\displaystyle{f{{\left({x}\right)}}}={2}{x}+{\cos{{x}}}$$
$$\displaystyle\text{Notice that: }\ {f{{\left(-{1}\right)}}}=-{2}+{\cos{{\left(-{1}\right)}}}{<}{0}$$
$$\displaystyle{f{{\left({1}\right)}}}={2}+{\cos{{\left({1}\right)}}}{>}{0}$$
Using the intermediate value theorem there exists c in (-1, 1) such that f(c) = 0.
This shows that f(x) has a root.
Now realize that:
$$\displaystyle{f}'{\left({x}\right)}={2}-{\sin{{\left({x}\right)}}}$$
Notice that $$\displaystyle{f}'{\left({x}\right)}{>}{0}$$ for all values of x.
Remember that Rolle's theorem states that if a function is continuous on [m, n] and differentiable on
(m, n) where f(m) = f(n) then there exists k in (m, n) such that f'(k) = 0.
Assume that this function has 2 roots :
$$\displaystyle{f{{\left({m}\right)}}}={f{{\left({n}\right)}}}={0}$$
Then there exists k in (m, n) such that f'(k) = 0.
Note however as I said:
$$\displaystyle{f}'{\left({x}\right)}={2}-{\sin{{\left({x}\right)}}}$$ is always positive so there exists no k such that f'(k) = 0.
This thus prove there cannot be two or more roots.
Hence:
$$\displaystyle{2}{x}+{\cos{{x}}}={0}$$
only has one real root.