Show analytically that 0 is the only zero of $\sin (2x)+2x$ (I know the question sounds a bit confusing but that's how it's written in my book.)
I tried:
$$0 = \sin(2x)+2x \Leftrightarrow \\
0 = 2\sin x \cos x + 2 x \Leftrightarrow \\
0 = 2(\sin x \cos x +x) \Leftrightarrow \\
0 = \sin x \sqrt{1-\sin^2x}+x \Leftrightarrow \\
0 = \sqrt{\sin^2x(1-\sin^2x)}+x \Leftrightarrow \\
0 = \sqrt{\sin^2x-\sin^4x}+x \Leftrightarrow \\
???
$$
What do I do next?
 A: Hint: There will be no zeros outside of the interval $[-1/2,1/2]$
A: Hint: if $f(x)=\sin(2x)+2x$ has more than one zero, then $f^{\prime}(x)$ has at least one zero by Rolle's theorem. Show that this is impossible in $[-1,1]$ (for instance).
A: Hint: Consider the increasing/decreasing behavior of the function $f(x) = \sin(2x) + 2x$
$f'(x) = 2\cos (2x) + 2$. Since the range of $\cos \theta$ is $[-1, 1]$, we know that $2\cos (2x) + 2 \ge 0$, so $f$ is (non-strictly) increasing for all real numbers.
A: Starting from $0 = \sin(2x)+2x$
Let $f(x)=\sin(2x) +2x$
$f'(x) = 2\cos(2x) + 2 \ge 0$
So it is non-decreasing (with a bunch of inflection points periodically where $f'(x)=0$)
Also notice that $f(0) = 0$ and $f'(0) > 0$, and thus $0$ is the only solution.
A: If we replace $2x$ by $x$ we see that we are looking for solutions
to $-\sin x = x$.
It is clear that we only need to consider $|x| \le 1$.
For $x \in [-1,0)$ we have $-\sin x > 0$ and $x <0$ and
for $x \in (0,1]$ we have $-\sin x < 0$ and $x >0$. Hence
the only point left ix $x=0$ which is the solution.
