Let X be a Banach space, \({\mathbb{C}}\) the complex numbers, and let f: \(X\to {\mathbb{C}}\) satisfy the functional equation \((A)\quad f(x+y)+(2f^ 2(y)- f(2y))f(x-y)=2f(x)f(y).\) (A) generalizes the well-studied equations of D'Alembert and Cauchy: \((D)\quad F(x+y)+F(x-y)=2F(x)F(y),\) and \((C)\quad G(x+y)=G(x)G(y),\) respectively. Assuming that f is not identically zero, substitution of 0 for y in (A) shows that either \(f(0)=1\) or \(f(0)=1/2\). In the latter case, 2f satisfies (C). Henceforth, we assume that \(f(0)=1\). Then (Theorem 2.2) there exist functions F and G from X to \({\mathbb{C}}\), satisfying (D) and (C) respectively, such that \(f(x)=G(x/2)F(x).\) This allows the classification of all solutions of (A), (Theorem 3.1): (i) f is continuous nowhere, (ii) f is continuous at some \(x_ 0\) where \(f(x_ 0)\neq 0\), or (iii) the set of points of continuity of f is nonempty and is contained in the set of zeros of f. In case (iii), f will be continuous precisely at its zeros, and the solution will be of the form \((S)\quad f(x)=\exp (A(x))\cosh (B(x))E(x),\) where A and B are linear functionals on X, B is continuous, and E: \(X\to {\mathbb{C}}\) satisfies (C) and \(| E(x)| =1\) for all x. In case (ii), f is again of the form (S), but with \(E\equiv 1\), and both A and B continuous linear functionals.