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The given equation is:

\[
\tan(\theta) + 1 = 0
\]

To solve this:

1. Subtract 1 from both sides:

\[
\tan(\theta) = -1
\]

2. Now, we need to find the angles \(\theta\) where the tangent of \(\theta\) is \(-1\) in the interval \([0, 2\pi]\). The general solution for \(\tan(\theta) = -1\) occurs where:

\[
\theta = \frac{3\pi}{4}, \frac{7\pi}{4}
\]

These are the two solutions for \(\theta\) in the interval \([0, 2\pi]\).

Thus, the solutions are:

\[
\theta = \frac{3\pi}{4}, \frac{7\pi}{4}
\]

Would you like further details or explanations?

Here are 5 related questions you might be interested in:
1. How do you derive the general solutions for trigonometric equations?
2. How do we determine solutions for \(\tan(\theta)\) in different quadrants?
3. What are the key characteristics of the tangent function in trigonometry?
4. How can you use reference angles to solve trigonometric equations?
5. What is the periodicity of the tangent function?

**Tip:** The tangent function has a period of \(\pi\), meaning its values repeat every \(\pi\) radians.

Find all solutions of the equation in the interval [0, 2π]. tan(θ) + 1 = 0.

Learn how to solve the trigonometric equation tan(θ) + 1 = 0 within the interval [0, 2π]. This problem explains key concepts of trigonometry, including the tangent function and its periodicity. Suitable for students in Grades 10-12.

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