https://file.aiquickdraw.com/mathbot/file/17290869079152b5xb8qk.jpg

The problem requires solving the equation:

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

for the interval \(-\pi \leq x < \pi\).

### Solution:
1. Start by isolating \(\tan(x)\):
   \[
   \tan(x) = -1
   \]

2. We need to find the values of \(x\) where \(\tan(x) = -1\) within the interval \(-\pi \leq x < \pi\).

3. The general solution for \(\tan(x) = -1\) is:
   \[
   x = \arctan(-1) + k\pi
   \]
   where \(k\) is an integer.

4. \(\arctan(-1)\) gives \(-\frac{\pi}{4}\), since \(\tan(-\frac{\pi}{4}) = -1\).

5. Applying the general solution:
   \[
   x = -\frac{\pi}{4} + k\pi
   \]

6. Now we check values of \(k\) that satisfy \(-\pi \leq x < \pi\):
   - For \(k = 0\):
     \[
     x = -\frac{\pi}{4}
     \]
     which is within the interval.
   
   - For \(k = 1\):
     \[
     x = -\frac{\pi}{4} + \pi = \frac{3\pi}{4}
     \]
     which is also within the interval.
   
   - For \(k = -1\):
     \[
     x = -\frac{\pi}{4} - \pi = -\frac{5\pi}{4}
     \]
     which is outside the interval.

   Thus, the only valid solutions within the interval \(-\pi \leq x < \pi\) are:
   \[
   x = -\frac{\pi}{4} \quad \text{and} \quad x = \frac{3\pi}{4}
   \]

### Final Answer:
The solutions are:
\[
x = -\frac{\pi}{4} \quad \text{and} \quad x = \frac{3\pi}{4}
\]

Would you like further details or have any questions on this solution?  

Here are 5 related questions for further exploration:
1. What is the general form of the solution for \(\tan(x) = 1\) over any interval?
2. How does the period of the tangent function influence the solutions to \(\tan(x) = c\)?
3. How would the solution change if the interval was \([0, 2\pi)\) instead?
4. What are the solutions to \(\tan(x) = \sqrt{3}\) in the interval \(-\pi \leq x < \pi\)?
5. How would you solve \(\sin(x) + \cos(x) = 0\) in the same interval?

**Tip:** The tangent function has a period of \(\pi\), so any solution to \(\tan(x) = c\) will repeat every \(\pi\) units along the x-axis.

Find the solution(s) to the equation tan(x) + 1 = 0 for -π ≤ x < π.

Learn how to solve the trigonometric equation tan(x) + 1 = 0 for -π ≤ x < π. This solution involves the concept of tangent function, the arctangent formula, and periodicity of trigonometric functions. Suitable for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation