Crazy Functional Equation - Possibly related to trig identities? $p(2x)=p(x)p(-x)-p(x-1)p(1-x)+4x-1$ I'm trying to find a set of nontrivial (identity function is not allowed, and neither are constant functions) continuous solutions to the functional equation
$$p(2x)=p(x)p(-x)-p(x-1)p(1-x)+4x-1$$
...and, to be honest, I'm not quite sure where to go with this. This functional equation looks vaguely trigonometric to me, so I tried making some functional substitutions, like
$$p(x)=\cos(\psi(x))$$
$$p(x)=\cos^2(\psi(x))$$
$$p(x)=\psi(x)\cos(x)$$
$$p(x)=\psi(x)\cos^2(x)$$
...but I got nowhere with that. Any ideas?

One can take trial functions, say 
$$p_{n}(x) = a_{0} + a_{1} \, x + a_{2} \, x^2 + \cdots + a_{n} \, x^n$$
and attempt to find the values of the coefficients. 
As an example consider the case of $n=2$, or $p_{2}(x) = a_{0} + a_{1} \, x + a_{2} \, x^2.$ By expanding terms one will find that
\begin{align}
a_{0} + 2 a_{1} x + 4 a_{2} x^2 &= (-1 - 3 a_{0} a_{2} + a_{1} a_{2} + a_{1}^2 - 2 a_{2}^2) + 2 \, x \, (2 + 2a_{0} a_{2} - a_{1}^2 + 2 a_{2}^2) \\
& \hspace{5mm} + x^2 \, (a_{0} - a_{1} - 5 a_{2}) \, a_{2} + 4 \, a_{2}^2 \, x^3 
\end{align}
Since $x^3$ only appears on one side of the equation then the coefficient is zero, or $a_{2} = 0$. This yields
 \begin{align}
a_{0} + 2 a_{1} x &= (-1 + a_{1}^2) + 2 \, x \, (2 - a_{1}^2). 
\end{align}
Now $a_{0} = a_{1}^2 - 1$ and $a_{1} = 2 - a_{1}^2$ which yields $a_{1} = \{1, -2\}$ and the set of values $(a_{0}, a_{1}) \in \{ (0,1), (3,-2)\}$. For $p_{2}(x)$ the two solutions are
$$p_{2}(x) = \begin{cases} x \\ 3- 2x \end{cases}.$$
Taking the case of $p_{1}(x) = a_{0} + a_{1} \, x$ then one finds that
$$p_{1}(x) = \begin{cases} x \\ 3- 2x \end{cases}.$$
Leaving the details for the interested it would be presumable to say that the solutions to this equation are:
Given the equation
$$p(2x)=p(x)p(-x)-p(x-1)p(1-x)+4x-1$$
then the solutions are
$$p(x) = \begin{cases} x \\ 3- 2x \end{cases}.$$
