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

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

Thanks in advance.