Cotθ+1=0. Find all solutions of the equation in the interval [0, 2π].

To solve the equation \(\cot \theta + 1 = 0\) within the interval \([0, 2\pi]\), we can start by isolating \(\cot \theta\):

\[
\cot \theta = -1
\]

The cotangent function, \(\cot \theta\), is the reciprocal of the tangent function:

\[
\cot \theta = \frac{1}{\tan \theta}
\]

Thus, when \(\cot \theta = -1\), it implies:

\[
\tan \theta = -1
\]

The tangent function \(\tan \theta\) is equal to \(-1\) at specific standard angles. In the unit circle, \(\tan \theta = -1\) when:

- \(\theta = \frac{3\pi}{4}\) (in the second quadrant)
- \(\theta = \frac{7\pi}{4}\) (in the fourth quadrant)

Both angles are in the specified interval \([0, 2\pi]\).

Therefore, the solutions to the equation \(\cot \theta + 1 = 0\) in the interval \([0, 2\pi]\) are:

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

If you have further questions or need more details, feel free to ask!

Here are some related questions for further exploration:
1. How can we find the general solution for \(\cot \theta = -1\)?
2. How does the cotangent function differ from the tangent function in terms of periodicity?
3. What are the key properties of the cotangent function on the unit circle?
4. How would you graph the equation \(\cot \theta = -1\) on the unit circle?
5. What is the significance of symmetry in solving trigonometric equations?
6. How do you determine the reference angle for \(\tan \theta = -1\)?
7. How can you verify the solutions using the unit circle?
8. How would you solve \(\cot \theta = 1\) instead?

**Tip:** Remember that trigonometric functions repeat their values in predictable patterns, so recognizing these can help you find all solutions within a given interval.

Learn how to solve the trigonometric equation \( \cot \theta + 1 = 0 \) within the interval [0, 2π]. Detailed step-by-step solution covering the properties of the cotangent function and its relationship with the tangent function. Suitable for high school students studying trigonometry.

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