Math Problem Statement

0 = tan + sec

Solution

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The equation appears to be:

0=tan(θ)+30 = \tan(\theta) + 3

To solve for θ\theta:

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

  2. Take the inverse tangent (arctan) of both sides to find θ\theta: θ=tan1(3)\theta = \tan^{-1}(-3)

  3. Using a calculator or in radians: θ1.249 radians\theta \approx -1.249 \text{ radians} In degrees, this would be: θ71.57\theta \approx -71.57^\circ

Since tangent is periodic, the general solution can be written as: θ=tan1(3)+nπ, where nZ\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!

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Tangent
Secant

Formulas

tan(θ) = opposite / adjacent
sec(θ) = 1 / cos(θ)

Theorems

Trigonometric Identities
Periodic Nature of Trigonometric Functions

Suitable Grade Level

Grades 10-12