Question

Solve 0=1-sin(2πT)

Answer 1

-1=-sin(2piT)
1=sin(2piT)

Answer 2

since we know 2pi is just a full revolution you can find T and it willhave the same value as 2piT

sin(t)=1

pi/2 is the only time sin = 1

Answer 3

Add 2piT both the sides,

\[\sin(2 \pi T) = 1\]

Now, for \[sinx = sina\]

The general solution is given by:

\[x = [n \pi + (-1)^n.a] , n \in Z\]

Answer 4

to get all values of sin=1 simply add 2pi n

\[\pi /2+2\pi n\]

Answer 5

really need to refresh my mind on trig identities

Answer 6

\[\sin(2 \pi T ) = 1\]

\[\sin(2 \pi T) = \sin \frac{\pi}{2}\]

\[2 \pi T = n \pi + (-1)^n \times \frac{\pi}{2}\]

\[T = \frac{n}{2} + (-1)^n \times \frac{1}{4} = \frac{n}{2} + \frac{1}{4}(-1)^n\]
So the general solution will be:

Answer 7

\[T = \frac{n}{2} + \frac{1}{4}(-1)^n\]