To solve the equation [tex]\(\tan \theta = -1\)[/tex], we start by recalling some fundamental properties of the tangent function and its periodicity. The tangent function [tex]\(\tan \theta\)[/tex] has a period of [tex]\(\pi\)[/tex], which means [tex]\(\tan(\theta + \pi) = \tan \theta\)[/tex] for all values of [tex]\(\theta\)[/tex].
Next, we identify the specific angles within one period [tex]\([0, 2\pi)\)[/tex] where [tex]\(\tan \theta = -1\)[/tex]. By examining the standard unit circle or tangent values, we find that:
- [tex]\(\tan(135^\circ) = \tan \left(\frac{3\pi}{4}\right) = -1\)[/tex]
- [tex]\(\tan(315^\circ) = \tan \left(\frac{7\pi}{4}\right) = -1\)[/tex]
Therefore, the two primary solutions in the interval [tex]\([0, 2\pi)\)[/tex] are [tex]\(\theta = \frac{3\pi}{4}\)[/tex] and [tex]\(\theta = \frac{7\pi}{4}\)[/tex].
However, since the tangent function is periodic with a period of [tex]\(\pi\)[/tex], we can express the general solutions for [tex]\(\theta\)[/tex] by adding integer multiples of the period [tex]\(\pi\)[/tex] to these specific solutions.
For [tex]\(\theta = \frac{3\pi}{4}\)[/tex]:
[tex]\[
\theta_1 = \frac{3\pi}{4} + k\pi \quad \text{where } k \text{ is any integer.}
\][/tex]
For [tex]\(\theta = \frac{7\pi}{4}\)[/tex]:
[tex]\[
\theta_2 = \frac{7\pi}{4} + k\pi \quad \text{where } k \text{ is any integer.}
\][/tex]
Hence, the general solutions to the equation [tex]\(\tan \theta = -1\)[/tex] are:
[tex]\[
\theta = \pi k + \frac{3\pi}{4} \quad \text{and} \quad \theta = \pi k + \frac{7\pi}{4}, \quad \text{where } k \text{ is any integer.}
\][/tex]
