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Consider the unit circle, where $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $.
For $ \tan(\theta) = -1 $, this implies that $ \sin(\theta) = -\cos(\theta) $.
Hence, $ \theta $ must be in the second or fourth quadrant, where sine and cosine have opposite signs.
This occurs at:

$ \theta = \frac{3\pi}{4} $

and

$ \theta = \frac{7\pi}{4} $

Therefore, the solutions to the equation $ \tan(\theta) = -1 $ in the interval $ [0, 2\pi) $ are:

$ \theta = \frac{3\pi}{4} \text{ and } \frac{7\pi}{4} $

Given $ an( heta) = -1 $,
and using the identity $ an( heta) = sin( heta) / cos( heta) $, we get $ sin( heta) = -cos( heta) $.
This occurs when $ heta $ is in the second and fourth quadrants.
In the second quadrant,

$ heta = pi – frac{pi}{4} = frac{3pi}{4} $

In the fourth quadrant,

$ heta = 2pi – frac{pi}{4} = frac{7pi}{4} $

Thus, the angles for which $ an( heta) = -1 $ are:

$ heta = frac{3pi}{4} ext{ and } frac{7pi}{4} $

To solve $ an( heta) = -1 $, we know that $ sin( heta) = -cos( heta) $.

This condition is satisfied in the second and fourth quadrants:

$ heta = frac{3pi}{4} $

and

$ heta = frac{7pi}{4} $

Hence, the angles are:

$ heta = frac{3pi}{4} ext{ and } frac{7pi}{4} $

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