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To solve for $\theta$ such that $\tan(\theta) = -1$ in the interval $[0°, 360°]$, we first recognize that the tangent function is negative in the second and fourth quadrants.

In the second quadrant, $\tan(180° – \theta) = -1$. So:

$180° – \theta = 45°$

Solving for $\theta$:

$\theta = 180° – 45°$

$\theta = 135°$

In the fourth quadrant, $\tan(360° – \theta) = -1$. So:

$360° – \theta = 45°$

Solving for $\theta$:

$\theta = 360° – 45°$

$\theta = 315°$

Thus, the values of $\theta$ that satisfy $\tan(\theta) = -1$ in the interval $[0°, 360°]$ are:

$\boxed{135°, 315°}$

We need to identify angles $ heta$ in $[0°, 360°]$ where $ an( heta) = -1$. The tangent function is negative in the second and fourth quadrants. Considering the unit circle, $ an( heta) = -1$ at:

$ heta = 180° – 45°$

$ heta = 135°$

and

$ heta = 360° – 45°$

$ heta = 315°$

Therefore, $ heta$ where $ an( heta) = -1$ within $[0°, 360°]$ are:

$oxed{135°, 315°}$

Solving $ an( heta) = -1$ in the interval $[0°, 360°]$, we get:

$ heta = 135° ext{ and } 315°$

Thus:

$oxed{135°, 315°}$

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