Function problem 5

We have $f(x)=e^x-cosx+x$ , then

$g(x)=e^x+sinx+1>0$ .

$f(x)$ is increasing function. So when we take limit where $x$ tends to $-\infty$ then $f(x) < 0$ and $f(x) > 0$ only if $x>0$ so $f(x) = 0$ has only $\boxed{1}$ root....