= Solution
Here $I(0)=1/2$. First restrict $\|u'\|_\infty\leq\pi/12$. On that slope interval, the second derivative of $h(p)=1-\cos(\pi/3+p)$ is $\cos(\pi/3+p)\geq\cos(5\pi/12)>1/4$. <Taylor's theorem> gives
$$
h(p)\geq h(0)+h'(0)p+\tfrac18p^2.
$$
The linear term integrates to zero because $\int_0^1u'=0$. Since $x^3\leq1$, the <Poincaré inequality> yields
$$
I(u)-I(0)\geq\frac18\int_0^1(u')^2-\int_0^1u^2
\geq\left(\frac18-\frac1{\pi^2}\right)\int_0^1(u')^2.
$$
The coefficient is positive. Thus \b[zero is a strict weak local minimizer].
For failure of strong minimality, take a triangular bump supported on $[1/2-\varepsilon,1/2+\varepsilon]$, with slopes $+2\pi$ and $-2\pi$, and zero outside. Its height is $2\pi\varepsilon\to0$. <Almost everywhere> its slope is $0$ or $\pm2\pi$, so the cosine term has exactly its zero-function value everywhere. The potential term is strictly negative on the bump. Therefore
$$
I(u_\varepsilon)-I(0)=-\int_0^1x^3u_\varepsilon^2\,dx<0,
\qquad\|u_\varepsilon\|_\infty\to0.
$$
Hence \b[it is not a strong local minimizer]. The derivatives of these bumps stay large, so they do not contradict weak local minimality.
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