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It looks like you've uploaded an image containing a trigonometric expression. I'll transcribe the equation and help you solve it.

The equation appears to be:

\[
0 = \tan(\theta) + 3
\]

To solve for \(\theta\):

1. Move the constant to the other side:
   \[
   \tan(\theta) = -3
   \]

2. Take the inverse tangent (arctan) of both sides to find \(\theta\):
   \[
   \theta = \tan^{-1}(-3)
   \]

3. Using a calculator or in radians:
   \[
   \theta \approx -1.249 \text{ radians}
   \]
   In degrees, this would be:
   \[
   \theta \approx -71.57^\circ
   \]

Since tangent is periodic, the general solution can be written as:
\[
\theta = \tan^{-1}(-3) + n\pi, \text{ where } n \in \mathbb{Z}
\]
This accounts for the periodic nature of the tangent function.

Would you like more details on solving trigonometric equations?

Here are 5 questions to consider next:
1. What is the periodicity of the tangent function?
2. How do you convert radians to degrees?
3. What is the importance of the general solution for trigonometric functions?
4. How can we solve other trigonometric equations involving sine and cosine?
5. What are the key differences between the tangent and cotangent functions?

**Tip:** Always remember to account for the periodicity of trigonometric functions when solving for all possible solutions!

This page covers the detailed steps to solve a trigonometric equation involving tangent and secant functions. Understand how to manipulate trigonometric identities to solve for θ in the equation 0 = tan + sec. This problem is ideal for high school students (Grades 10-12) studying advanced trigonometry.

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