Solution
= Solution
$\sin x=\sum_{n\ge0}(-1)^nx^{2n+1}/(2n+1)!$ and $\cos x=\sum_{n\ge0}(-1)^nx^{2n}/(2n)!$. Here $\sin\prime\prime\prime=-\cos$ is negative while $\cos\gt 0$ near zero; applying the preceding shape argument until the first zero of $\cos$ and then the addition identities establishes a first positive zero $x_0$ with positivity before it.