= Solution
Let $u,w\in M$ both solve the <variational inequality>. Test the inequality for $u$ with $w$ and the one for $w$ with $u$, then add. Writing $d=u-w$ gives
$$
2\int_0^la|d'|^2\,dx\leq-\gamma\int_0^l(\cos u-\cos w)d\,dx\leq\gamma\|d\|_2^2,
$$
since <cosine> is <Lipschitz> with constant one. Since $d(0)=0$, the <fundamental theorem of calculus> and <Cauchy-Schwarz inequality> give $|d(x)|\leq\sqrt l\,\|d'\|_2$; integrating gives $\|d\|_2^2\leq l^2\|d'\|_2^2$. Therefore
$$
(2c_0-\gamma l^2)\|d'\|_2^2\leq0.
$$
For $\boxed{0<\gamma<2c_0/l^2}$, the coefficient is positive, so $d'=0$ and the zero boundary value gives $d=0$. This proves uniqueness with an explicit sufficient smallness condition. More generally any <Poincaré inequality> constant $C_P$ gives the condition $\gamma C_P^2<2c_0$; optimizing that constant is unnecessary for the requested conclusion. The underlying principle is <uniqueness for a strongly monotone variational inequality>.
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