Let $\displaystyle \phi_{1}$ and $\displaystyle \phi_{2}$ be two solutions on $\displaystyle \mathbb{R}$ of the differential equation
$\displaystyle y^{''} + sin^{2}(x)y = 0$,
such that $\displaystyle \phi_{1}(0) = 1$, $\displaystyle \phi^{'}_{1}(0) = 0$ and $\displaystyle \phi_{2}(0) = 0$, $\displaystyle \phi^{'}_{2}(0) = 1$. Let $\displaystyle \phi(x)$ be the function defined by
$\displaystyle \phi(x) = \phi_{2}(\pi)\phi_{1}(x) + (1 - \phi_{1}(\pi))\phi_{2}(x)$.
(a) Show that $\displaystyle \phi$ is also a solution of $\displaystyle y^{''} + sin^{2}(x)y = 0$.
(b) Suppose $\displaystyle \phi_{1}(\pi) + \phi^{'}_{2}(\pi) = 2$. Show that $\displaystyle \phi(x) = \phi(x + \pi)$ for all $\displaystyle x \in \mathbb{R}$.

I've managed to do part (a) and the part $\displaystyle \phi(0) = \phi(\pi)$. However, I have difficulty in proving that $\displaystyle \phi^{'}(0)=\phi^{'}(\pi)$ using the condition stated in part (b) in order to show that $\displaystyle \phi(x) = \phi(x + \pi)$ using the existence and uniqueness theorem. Could anyone help with this?

Thanks in advance.