How to solve this spring (mechanical vibrations) problem

A body with mass 250g is attached to the end of a spring. (It takes 9 N to stretch the spring 25 cm) At time t=0, the body is pulled 1 m to the right, stretching the spring, and set in motion with an initial velocity of 5m/s to the left. Fin the position function of the body.

The answer is x(t)=(13/12)cos(12t - 5.8884) but I have no idea how to get there. Any help is aprreciated.

Find the spring constant:
F = k × x
(9 N) = k × (0.25 m)
k = 36 N/m

Find the the angular velocity of the body:
ω = √( k / m )
ω = √( (36 N/m) / (0.25 kg) )
ω = 12 rad/s

Find the amplitude of the SHM.
The elastic potential energy at t=0 is:
Ee = k × x² / 2
Ee = (36 N/m) × (1 m)² / 2
Ee = 18 J
The extra kinetic energy given to the mass at t=0 is:
Ek = m × v² / 2
Ek = (0.25 kg) × (-5 m/s)² / 2
Ek = 3.125 J
The total mechanical energy at t=0 is:
Em = Ee + Ek
Em = (18 J) + (3.125 J)
Em = 21.125 J
So the amplitude of the SHM is:
Em = k × A² / 2
21.125 J = (36 N/m) × A² / 2
A = 13/12 m

The general equation for position of SHM is
x(t) = A×cos(ω×t + φ)

You already know A and ω, and you know that x at t=0 is 1 m, so
(1 m) = (13/12 m) cos((12 rad/s)×(0 s) + φ)
(12/13 m) = cos(φ)
(put your calculator in RAD modus)
φ = 0.3948 rad
OR
φ = (2π - 0.3948) rad = 5.8884 rad

Which one of those values is the correct one, can be determined by finding the equation for velocity
The equation for velocity is the first derivative of the equation for position, so
v(t) = -A×ω×sin(ω×t + φ)
v(t) = -(13/12 m)×(12 rad/s)×sin((12 rad/s)×t + φ)
v(t) = -(13 m/s)×sin((12 rad/s)×t + φ)

Now try using φ = 0.3948 rad at t=0
v(t) = -(13 m/s)×sin((12 rad/s)×(0 s) + (0.3948 rad))
v(t) = -(13 m/s)×sin(0.3948)
v(t) = -5 m/s

And try using φ = 5.8884 rad at t=0
v(t) = -(13 m/s)×sin((12 rad/s)×(0 s) + (5.8884 rad))
v(t) = -(13 m/s)×sin(5.8884)