General Solution of Differential Equation with Trigonometric Functions

Question

I'm trying to solve the differential equation $x'' + 4x = 4t$, endpoint value $x(0) = x(1) = 0$, and ask me how to derive the solution $x(t) = t - \sin2t/\sin2$.

The answer says that since the particular solution $t$ is easily found, the general solution is $A\cos2t + B\sin2t + t$.

But I think that due to the characteristic equation being $r^2 - 4 = 0$ the general solution should be $Ae^{2t} + Be^{-2t} + t$.

Why can this be written in terms of trigonometric functions?

Thanks in advance.

Answer 1

Here, the characteristic equation is $r^2 + 4=0$.