$x = c_1 \cos \omega t + c_2 \sin \omega t$ ← equation (1)
$\dfrac{dx}{dt} = -\omega c_1 \, \sin \omega t + \omega c_2 \, \cos \omega t$
$\dfrac{d^2x}{dt^2} = -\omega^2 c_1 \, \cos \omega t - \omega^2 c_2 \, \sin \omega t$
$\dfrac{d^2x}{dt^2} = -\omega^2 (c_1 \, \cos \omega t + c_2 \, \sin \omega t)$
From equation (1), c1 cos ωt + c2 sin ωt = x
$\dfrac{d^2x}{dt^2} = -\omega^2 x$
$\dfrac{d^2x}{dt^2} + \omega^2 x = 0$ answer