Question

Find all solutions between 0 and 2pi. Round to two decimal places. In radians.
Find all solutions between 0 and 2 pi. Round to two decimal places for the final solutions. The answers should be in radian mode. If you can use exact values use them. At least one of the 3 problems can use exact values for the answers and work to get to the answers.

2. 150sin3s=27


3. 16cos(3t-1)+4=-8



4. 9sin(pi/2(t-1))+5=-1

Answer 1

2.
sin 3s = 27/150
3s = .18198 in I or 3s = 2.96 in II
s = .06 or s = .99
but period of sin 3s = 2π/3
so by adding 2π/3 to any existing answer yields a new answer

x = .06 , 2.15 , 4.25
x = .99 , 3.08, 5.18

3.
16 cos(3t-1) = -12
cos(3t-1) = -12/16 = -.75
3t - 1 is in II or III
3t-1 = 2.42 or 3t-1 = 3.86
3t = 3.42 or 3t = 4.86
t = 1.14 or t = 1.62
again, period of cos(3t-1) = 2π/3

t = 1.14 , 3.23 , 5.33
t = 1.62 , 3.72 , 5.81

4.
9sin(pi/2(t-1)) = -6
sin(pi/2(t-1)) = -6/9 = -2/3
(π/2)(t-1) = 3.871 or (π/2)(t-1) = 5.55

t-1 = 2.46 or t = 3.54
t = 3.46 or t = 4.54
period = 2π/(π/2) = 4
but by adding 4 we go beyond the domain

so t = 3.46 or t = 4.54

Answer 2

Use the graph of the sine function to find all the solutions of the equation. (Enter your answer in the form a + bπn, where a [0, 2π), b is the smallest possible positive number, and n represents any integer.)
cos(t) = -1

I've tried everything to understand the problem. I have gotten to a point where I can get the correct answer but I do not understand how to put it in the form that it wants.