Math Problem Statement

Solve each equation for the given interval. First equation: 3 tan θ - 3 = 5 tan θ - 1, where -π ≤ θ ≤ π. Second equation: 2 csc θ = 6, where -180° ≤ θ ≤ 90°.

Solution

Let's start by solving the two problems from the image you've provided:

Problem 1 (Solve for πθπ-\pi \leq \theta \leq \pi):

Equation:

3tanθ3=5tanθ13 \tan \theta - 3 = 5 \tan \theta - 1

Steps:

  1. Move all terms involving tanθ\tan \theta to one side: 3tanθ5tanθ=1+33 \tan \theta - 5 \tan \theta = -1 + 3 2tanθ=2-2 \tan \theta = 2

  2. Solve for tanθ\tan \theta: tanθ=22=1\tan \theta = \frac{2}{-2} = -1

  3. Find θ\theta such that tanθ=1\tan \theta = -1 in the interval πθπ-\pi \leq \theta \leq \pi:

    • The principal solution is θ=π4\theta = -\frac{\pi}{4}, where tanθ=1\tan \theta = -1.
    • The general solution for tanθ=1\tan \theta = -1 is: θ=π4+nπfor any integer n.\theta = -\frac{\pi}{4} + n\pi \quad \text{for any integer } n. Considering the given interval πθπ-\pi \leq \theta \leq \pi, the solutions are: θ=π4,θ=3π4.\theta = -\frac{\pi}{4}, \quad \theta = \frac{3\pi}{4}.

Thus, the solutions are: θ=π4,3π4.\boxed{\theta = -\frac{\pi}{4}, \frac{3\pi}{4}}.


Problem 2 (Solve for 180θ90-180^\circ \leq \theta \leq 90^\circ):

Equation:

2cscθ=62 \csc \theta = 6

Steps:

  1. Solve for cscθ\csc \theta: cscθ=62=3\csc \theta = \frac{6}{2} = 3

  2. Recall that cscθ=1sinθ\csc \theta = \frac{1}{\sin \theta}, so: sinθ=13\sin \theta = \frac{1}{3}

  3. Find θ\theta such that sinθ=13\sin \theta = \frac{1}{3} in the interval 180θ90-180^\circ \leq \theta \leq 90^\circ.

    • The principal solution is θ=arcsin(13)19.47\theta = \arcsin\left(\frac{1}{3}\right) \approx 19.47^\circ.
    • Since the sine function is positive in the first and second quadrants, the solution for the second quadrant within the given interval is θ=18019.47=160.53\theta = 180^\circ - 19.47^\circ = 160.53^\circ, but this is outside the given range, so we discard it.

Therefore, the only solution is: θ19.47.\boxed{\theta \approx 19.47^\circ}.


Would you like further details on any of the steps? Here are 5 follow-up questions that relate to this content:

  1. How can you derive the general solution for a trigonometric equation?
  2. What is the significance of the reference angle when solving for trigonometric functions?
  3. How does the periodicity of the tangent function affect the number of solutions in a given interval?
  4. Can you explain how the sine function behaves in different quadrants?
  5. How do you convert angles between radians and degrees?

Tip: When solving trigonometric equations, always consider the behavior of the function in each quadrant and any restrictions on the angle.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Solving Trigonometric Equations
Tangent Function
Cosecant Function

Formulas

tan θ = sin θ / cos θ
csc θ = 1 / sin θ
General solution for tan θ = -1

Theorems

Tangent and Cosecant Function Periodicity
Unit Circle Reference

Suitable Grade Level

Grades 10-12