Answer: $a \in \mathbb{R} \setminus \{1\}$.

First, if $a \neq 1$ then it is easy to verify that the function
$$f(x) = \frac{2x^2 - 1}{a - 1}$$
satisfies the given equation.

On the other hand, for $a = 1$ we have the functional equation
$$f(\sin x) + f(\cos x) = \cos 2x,$$
which after changing $x$ by $\pi/2 - x$ becomes
$$f(\cos x) + f(\sin x) = - \cos 2x,$$
i.e. $\cos 2x = - \cos 2x$ for all real $x$, a contradiction.