Why there are only $2$ roots for $f(x) = x^4 + x^2 + \sin x$? Why there are only $2$ roots for $0 = x^4 + x^2 + \sin x$ ? It is obvious that $x = 0$ is one root but I couldn't prove that there is only one more root for this function
 A: Suppose that $f$ has three real roots. Then by Rolle, $f'$ has two real roots. Again by Rolle: $f'' $ has a real root. But $f'' \ge 1$ on $ \mathbb R$.
A: Really you have four roots (apparently). Two real and two complex (?) Making 
$$
f(z) = z^4+z^2+\sin z = 0,\;\; z = x + i y
$$
is equivalent after expansion $(f(z) = u(x,y)+i v(x,y))\;\;$ to the set of real functions
$$
u(x,y) = x^4+y^4+x^2-y^2-6x^2y^2-\cosh y \sin x = 0\\
v(x,y) = 4x^3y-4x y^3+2 x y+\sinh y\cos x = 0
$$
Follows a first plot showing in blue $u(x,y)=0\;\;$ and in red $v(x,y) = 0\;$ Their intersections (four) two in the real axis, and two in the complex plane, are the four (?) roots.

but surprisingly widening the plot scope we can appreciate a lot (infinite) more of complex roots. 

Concluding. Two real roots and infinite more complex.
A: Look at the 2nd derivative; it's always positive:
$$f''(x) = 12x^2 + 2 - \sin x \geq 0 + 2 - 1 = 1$$
Since $f \in C^2(\mathbb R),$ the positiveness of the 2nd derivative implies that there can be no more than two roots.
Furthermore, since $f(0) = 0,$ $f'(0) > 0$ and $\lim_{x \to -\infty} f(x) = +\infty,$ there must be another root to the left of $0.$
Thus there are exactly two roots.
A: $f''(x)>0$, $f$ is convex, at most $2$ real roots .
1) $f(0)= 0$ .
2) Let $x<0$:
$\lim_{x \rightarrow -\infty}f(x)=+\infty.$
$f(-π/6)=$
$(π/6)^4+(π/6)^2 - 1/2<0.$
Since $f$ is continuos there is a zero in $(-\infty, -π/6)$.
