Polynomial equation: $P(\sin t) = P(\cos t)$

From $P(x)=P(\sqrt{1-x^2})$ we conclude $P$ is comprised of even powers, so $P(x)=L(x^2)$.

Now the condition reads $L(u)=L(1-u)$. Divide $L(u)$ by $u(1-u)$ to obtain quotient $q(u)$ and remainder $r(u)$. Substitute $L(u)=q(u)\,u(1-u)+r(u)$ into $L(u)=L(1-u)$ then reduce modulo $u(1-u)$ to obtain $au+b\equiv a(1-u)+b$, hence $a=0$. Then the functional equation descends to $q(u)=q(1-u)$ for the quotient $q$. Induct on degree to get $L(u)=Q(u(1-u))$.

Therefore $P(x)=L(x^2)=Q(x^2(1-x^2))$.