”Determine
Answer 1
Chloe Evans
To determine the general solution for $ \sin(x) = \frac{1}{2} $ within the interval $ [0, 2\pi] $, we need to find all the angles $ x $ where the sine function yields $ \frac{1}{2} $ on the unit circle.
The sine function equals $ \frac{1}{2} $ at angles $ \frac{\pi}{6} $ and $ \frac{5\pi}{6} $ within the given interval.
Thus, the general solutions are:
$ x = \frac{\pi}{6} + 2n\pi $
and
$ x = \frac{5\pi}{6} + 2n\pi $
where $ n $ is any integer.
Answer 2
Mia Harris
To find all solutions to $ sin(x) = -frac{1}{2} $ for $ x $ in the interval $ [0, 2pi] $, we look for angles $ x $ on the unit circle where the sine value is $ -frac{1}{2} $.
The sine function equals $ -frac{1}{2} $ at angles $ frac{7pi}{6} $ and $ frac{11pi}{6} $ within the given interval.
So, the solutions are:
$ x = frac{7pi}{6} $
and
$ x = frac{11pi}{6} $
Answer 3
Henry Green
To solve $ sin(x) = -1 $ within the interval $ [0, 2pi] $, we identify the angle $ x $ where the sine value is $ -1 $.
The sine function equals $ -1 $ at angle:
$ x = frac{3pi}{2} $
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