How do you solve $sin theta= 1$ ?

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Answer 1 · 2017-01-16T10:26:36

$theta=pi/2+-n2pi$ where n is a natural number

Let's look at a graph of $sintheta$ :

graph{sinx [-6.25, 6.25, -2, 2]}

This is $sintheta$ with $-2pi < theta < 2pi$

We can see that $sintheta=1$ at $pi/2$ and will repeat each cycle of the graph. And so the answer is:

$theta=pi/2+-n2pi$ where n is a natural number

How do you solve sin theta= 1sin theta= 1?