Help me with this trig problem

Okay so ive done this question and the last few lines of the solution is :

sin (pi/6)t = -1

pi/6 t = 3pi/2

t = 3pi/2 x 6/pi

t= 9

Can someone explain how i get from the 1st line to the 2nd line because i dont understand where the sine disappeared to.

thanks :)

sin (π /6)t = -1

The value of sine x = -1 is sin (3π /2) = -1

Then u write like this :

sin (π /6)t = sin (3π /2)

π /6 t = 3π /2

t = 3π/2 x 6/π

t= 9

sin (pi/6)t = -1
In the RHS you have (- 1)
Ask the question : what is that angle whose sin is (-1)?
what answer do you get ? obviously 3pi/2
So sin (pi/6)t = sin (3pi/2) => (pi/6)t = 3pi/2
=> t / 6 = 3 / 2 => t = 6 * (3 / 2) = 18/2 = 9

As written t = -2:
sin (pi/6)*t = -1
1/2 * t = -1
t = -2

But perhaps you meant:
sin[(pi*t)/6] = -1
[(pi*t)/6] = arcsin(-1)
[(pi*t)/6] = -pi/2 + 2kpi
[(pi*t)/6] = 3pi/2 + 2kpi
[(pi*t)/6] = 3pi/2
2 pi t = 18 pi
t = (18 pi) /(2 pi) = 9

The value of the sin must be equal to: -1/3pi2.

apply arcsine on both sides