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.