How to find the equation of the trajectory of a body thrown at an angle of 30 degrees

How to find the equation of the trajectory of a body thrown at an angle of 30 degrees to the horizon with an initial velocity of 3 m / s?

Displacement projections on the coordinate axes:
x = v0 * cosα * t
y = vo * sinα * t-g * t² / 2
Let us express the coordinate x in terms of time and substitute it into the equation for y:
t = x / v0 * cosα
y = vo * sinα * x / v0 * cosα – g * (x / v0 * cosα) ² / 2
y = x * tgα-g * x² / (2 * v0²cosα²)
Let’s substitute our data:
y = x * tgα-g * x² / (2 * v0²cosα²) = x * √3 / 3-x² * 0.72
substitute for x v0 * cosα * t:
y = x * √3 / 3-x² * 0.72 = (√3 / 3) * v0 * cosα * t – 0.72 * (v0 * cosα * t) ² = 1.5 * t – 4.86 t²
Answer: y = 1.5 * t – 4.86 t².