simple harmonic motion

The equation x=Asin(ωt+ψ) is a general solution for a simple harmonic motion. It is found by solving the second order deifferential equation describing the motion.

The values of ψ, the phase angle, indeed, depends on the "initial condition" of the motion ("initial condition" means the displacement of the mass at t = 0). If the oscillating mass starts at the equilibrium position (i.e. at t= 0 ), because we would expect x = 0 when t = 0, ψ should take the value of zero. The general equation thus reduces to x=Asin(ωt).

On the other hand, if the mass starts at a +ve max displacement, we would expect x = A when t = 0, then ψ should be equal to pi/2 radians (or 90 degrees). Because x=Asin(ωt+pi/2) can be written as x=Acos(ωt)

Similarly, if the mass starts at -ve max displacement, we would expect x = -A when t = 0. Under that situation, ψ should be equal to -pi/2 radians. Because x=Asin(ωt-pi/2) can be written as x= -Acos(ωt)

In your given problem, it is clear that the phase angle ψ = pi/2 radians, so that the initial condition of x = +2 cm at t = 0 s can be satisfied.
In such case, the equation becomes, x=(2)sin(ωt+pi/2) = (2)cos(ωt)

Actually, both x=Asin(ωt+ψ) and x=Acos ωt are solutions to the second order differential equations describing the motion of object under s.h.m. The reason why using sine or cosine is that cyclic properties of the motions are observed. And x=Asin(ωt+ψ) is the general solution s.h.m. ψ, being the initial phase angle, describes the initial state of the motion. And if
ψ = pi / 2, Asin(ωt+ψ) becomes Acos ωt. Generally, we use Asin(ωt+ψ) to describe motion since the phase angle is involved.

As to your example, both expressions can be used to solve the problem.
First, let x = Asin(ωt+ψ), A = amplitude
In the example, A = 1 (cm), so that at t = 0 (s),
sin(ψ) = 1 and cos(ψ) = 0
And at t = t0(s) = 0.03 (s), x = 0(cm), as it is at equilibrium position, so that
sin(ωt0+ψ) = 0, and knowing that sin(A+B) = sinAcosB + cosAsinB
Then, sin(ωt0)cos(ψ) + cos(ωt0)sin(ψ) = 0
cos(ωt0) = 0 and ωt0 = pi / 2, ω = pi/ (2t0)
Now, A = 2 (cm), and let at time t1, the mass rises 1 (cm) so that
2sin(ωt1+ψ) = 1
sin(ωt1+ψ) = 1/2
Expanding, cos(ωt1) = 1/2
t1 = pi / (3ω)
Substituing and noting that t0 = 0.3 (s), t1 = 0.2 (s)
You can find the solution similarly by using the other expression. And this time, using cosine is quicker as the initial position of the mass is equal to the amplitude of the motion.