Last trigonometry homework... HELP ASAP?

Question

2. a weight is suspended from a spring and vibrating vertically according to the equation y=2sin4pi(T+1/8) where y cm is the directed distance of the weight from its central position T seconds after the start of the motion and positive direction is upward

a) solve the equation for T

b) use the equation to determine the smallest 3 positive values of T for which the weight is 1cm above the central position.

please help me with it...I'm not that good in trigonometry...

Answer 1

Part a)

y/2 = sin [ 4π (T + 1/8) ]

sin^(-1) (y/2) = 4πT + π/2

4πT = sin^(-1) (y/2) - π/2

T = [ sin^(-1) (y/2) - π/2 ] / (4π)

Part b)

y = 1

T = [ sin^(-1) (1/2) - π/2 ] / (4π)

T1 = [ 5π/6 - π/2 ] / (4π) = 1 / 12

T2 = [ 13π/6 - π/2 ] / (4π) = 5 / 12

T3 =  [ 17π/6 - π/2 ] / (4π) = 7 / 12

Answer 2

a)

y = 2 sin [4pi (T + 1/8)]

y/2 = sin [4piT + pi/2]

sin^(-1) [y/2] = 4piT + pi/2

-(pi/2) + sin^(-1) [y/2] = 4piT

(1 / 4pi)[(-pi/2) + sin^(-1) [y/2]] = T

b)

y = 1, so...

T = (1 / 4pi)[(-pi/2) + sin^(-1) [1/2]]

sin^(-1) (1/2) = pi/6, 5pi/6, 7pi/6

So....

if sin^(-1) 1/2 = pi/6

T = (1 / 4pi)[(-pi/2) + (pi/6)]

= (1 / 4pi)(-3pi/6 + pi/6)

= (1/4pi)(-2pi/6)

= -2pi / 24pi

= -1/12

Do the same for if

sin^(-1) 1/2 = 5pi/6 and

sin^(-1) 1/2 = 7pi/6