Math Problem Statement

14tan(pitheta-4)+31=10

Solution

We need to solve the equation:

14tan(πθ4)+31=1014 \tan(\pi \theta - 4) + 31 = 10

Steps:

  1. Isolate the trigonometric term: 14tan(πθ4)=103114 \tan(\pi \theta - 4) = 10 - 31 14tan(πθ4)=2114 \tan(\pi \theta - 4) = -21 tan(πθ4)=2114\tan(\pi \theta - 4) = \frac{-21}{14} tan(πθ4)=32\tan(\pi \theta - 4) = -\frac{3}{2}

  2. Solve for πθ4\pi \theta - 4: Recall that tan(x)=y\tan(x) = y implies: x=tan1(y)+nπ,nZx = \tan^{-1}(y) + n\pi, \quad n \in \mathbb{Z} Applying this to our equation: πθ4=tan1(32)+nπ,nZ\pi \theta - 4 = \tan^{-1}\left(-\frac{3}{2}\right) + n\pi, \quad n \in \mathbb{Z}

  3. Calculate tan1(3/2)\tan^{-1}(-3/2): Using approximate values or a calculator: tan1(32)0.9828radians.\tan^{-1}\left(-\frac{3}{2}\right) \approx -0.9828 \, \text{radians}.

    Substituting: πθ4=0.9828+nπ\pi \theta - 4 = -0.9828 + n\pi

  4. Solve for θ\theta: πθ=40.9828+nπ\pi \theta = 4 - 0.9828 + n\pi πθ=3.0172+nπ\pi \theta = 3.0172 + n\pi Divide through by π\pi: θ=3.0172π+n\theta = \frac{3.0172}{\pi} + n

    Approximating π3.1416\pi \approx 3.1416: θ=3.01723.1416+n\theta = \frac{3.0172}{3.1416} + n θ0.9602+n\theta \approx 0.9602 + n

General Solution:

θ=0.9602+n,nZ.\theta = 0.9602 + n, \quad n \in \mathbb{Z}.

Would you like a more detailed step-by-step breakdown or a specific evaluation for nn?


Follow-up Questions:

  1. What values of nn are most relevant to your problem?
  2. Should the solution be restricted to a specific interval, such as [0,1][0, 1]?
  3. Would you like the graph of this function to visualize the solutions?
  4. Are there additional trigonometric equations you'd like solved?
  5. Do you want to verify this solution with numerical methods?

Tip:

When working with trigonometric equations, always ensure you consider the periodicity of the functions to capture all possible solutions!

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Periodic Functions
Inverse Trigonometric Functions

Formulas

tan(x) = y implies x = arctan(y) + nπ, n ∈ Z
General solution for trigonometric equations with tangent: x = arctan(y) + nπ
Conversion between radians and fractions of π: θ = value/π

Theorems

Periodicity of Tangent Function
Inverse Tangent Function Theorem

Suitable Grade Level

Grades 10-12