Question: Find all solutions to the equation.
tan $\theta$ + $\sqrt{3}$ = 0
Write your answer in radians in terms of $\pi$, and use the "or" button as necessary.
Example: $\theta = \frac{\pi}{5} + 2k\pi$, $k\in\mathbb{Z}$ or $\theta = \frac{\pi}{7} + k\pi$, $k\in\mathbb{Z}$
$\theta =$
1. **State the problem:** We need to find all values of $\theta$ such that
$$\tan \theta + \sqrt{3} = 0$$
2. **Rewrite the equation:**
$$\tan \theta = -\sqrt{3}$$
3. **Recall the tangent values:**
We know that $\tan \frac{\pi}{3} = \sqrt{3}$.
Since tangent is negative, $\tan \theta = -\sqrt{3}$ means $\theta$ is an angle where tangent is negative and has magnitude $\sqrt{3}$.
4. **General solution for tangent:**
The tangent function has period $\pi$, so solutions are of the form
$$\theta = \alpha + k\pi, \quad k \in \mathbb{Z}$$
where $\alpha$ is a reference angle.
5. **Find reference angles:**
Since $\tan \frac{\pi}{3} = \sqrt{3}$, the angles where $\tan \theta = -\sqrt{3}$ are those where $\theta$ is in the second or fourth quadrants (where tangent is negative).
These angles are:
$$\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$
and
$$\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$$
6. **Write the general solutions:**
Since tangent has period $\pi$, the general solutions can be written as
$$\theta = \frac{2\pi}{3} + k\pi, \quad k \in \mathbb{Z}$$
or equivalently
$$\theta = \frac{5\pi}{3} + k\pi, \quad k \in \mathbb{Z}$$
7. **Final answer:**
$$\boxed{\theta = \frac{2\pi}{3} + k\pi, \quad k \in \mathbb{Z} \quad \text{or} \quad \theta = \frac{5\pi}{3} + k\pi, \quad k \in \mathbb{Z}}$$