Solving Trigonometric Equations

Overview

Solving trigonometric equations involves finding the values of angles that satisfy trigonometric functions within specific intervals. This section will introduce techniques for solving both basic and advanced trigonometric equations, using identities and inverse functions to simplify and find solutions.

Key Concepts

  • Basic Trigonometric Equations: Equations such as \( \sin(x) = \frac{1}{2} \), \( \cos(x) = 0 \), and \( \tan(x) = \sqrt{3} \).
  • Inverse Trigonometric Functions: Using functions like \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \) to find specific angle values.
  • Using Identities: Applying trigonometric identities to simplify complex equations and make them solvable, such as using Pythagorean identities, double angle identities, and sum/difference identities.
  • Solving Equations Over Specific Intervals: Ensuring that solutions are found for angles within a specific interval, usually \( [0, 2\pi) \) or \( [0^\circ, 360^\circ) \).

Techniques for Solving Trigonometric Equations

To solve trigonometric equations, follow these general steps:

  • Step 1: Isolate the trigonometric function.
  • Step 2: Use inverse trigonometric functions to find the principal solution.
  • Step 3: Find other solutions by considering the periodicity of the trigonometric function.
  • Step 4: Solve for angles within the specified interval.

Practice Problems

  1. Solve \( \sin(x) = \frac{1}{2} \) for \( x \in [0^\circ, 360^\circ) \).
    Solution

    We know that \( \sin(30^\circ) = \frac{1}{2} \). Thus, \( x = 30^\circ \). Since sine has a positive value in both the first and second quadrants, we also have \( x = 180^\circ - 30^\circ = 150^\circ \). Therefore, the solutions are \( x = 30^\circ \) and \( x = 150^\circ \).

  2. Solve \( \cos(x) = 0 \) for \( x \in [0^\circ, 360^\circ) \).
    Solution

    We know that \( \cos(90^\circ) = 0 \) and \( \cos(270^\circ) = 0 \). Therefore, the solutions are \( x = 90^\circ \) and \( x = 270^\circ \).

  3. Solve \( \tan(x) = \sqrt{3} \) for \( x \in [0^\circ, 360^\circ) \).
    Solution

    We know that \( \tan(60^\circ) = \sqrt{3} \), so one solution is \( x = 60^\circ \). Since the tangent function has a period of \( 180^\circ \), the other solution is \( x = 60^\circ + 180^\circ = 240^\circ \). Therefore, the solutions are \( x = 60^\circ \) and \( x = 240^\circ \).

  4. Solve \( 2\sin(x) - 1 = 0 \) for \( x \in [0^\circ, 360^\circ) \).
    Solution

    First, isolate the sine function: \( 2\sin(x) = 1 \), so \( \sin(x) = \frac{1}{2} \). We know that \( \sin(30^\circ) = \frac{1}{2} \), so one solution is \( x = 30^\circ \). Since sine is positive in the first and second quadrants, the other solution is \( x = 180^\circ - 30^\circ = 150^\circ \). Therefore, the solutions are \( x = 30^\circ \) and \( x = 150^\circ \).

  5. Solve \( \cos(2x) = \frac{1}{2} \) for \( x \in [0^\circ, 360^\circ) \).
    Solution

    We know that \( \cos(60^\circ) = \frac{1}{2} \). Therefore, \( 2x = 60^\circ \) or \( 2x = 360^\circ - 60^\circ = 300^\circ \). So, \( x = 30^\circ \) or \( x = 150^\circ \). Since the cosine function has a period of \( 360^\circ \), we must also consider \( x = 30^\circ + 180^\circ = 210^\circ \) and \( x = 150^\circ + 180^\circ = 330^\circ \). Therefore, the solutions are \( x = 30^\circ, 150^\circ, 210^\circ, 330^\circ \).

  6. Solve \( \tan(x) = -1 \) for \( x \in [0^\circ, 360^\circ) \).
    Solution

    We know that \( \tan(45^\circ) = 1 \), so \( \tan(225^\circ) = -1 \) and \( \tan(135^\circ) = -1 \). Therefore, the solutions are \( x = 135^\circ \) and \( x = 225^\circ \).

  7. Solve \( \sin(2x) = \frac{\sqrt{2}}{2} \) for \( x \in [0^\circ, 360^\circ) \).
    Solution

    We know that \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \), so \( 2x = 45^\circ \) or \( 2x = 180^\circ - 45^\circ = 135^\circ \). Thus, \( x = 22.5^\circ \) or \( x = 67.5^\circ \). Since sine has a period of \( 360^\circ \), we also have \( 2x = 45^\circ + 360^\circ = 405^\circ \), so \( x = 202.5^\circ \), and similarly for other solutions. The solutions are \( x = 22.5^\circ, 67.5^\circ, 157.5^\circ, 202.5^\circ \).

  8. Solve \( \cos(x) = -\frac{1}{2} \) for \( x \in [0^\circ, 360^\circ) \).
    Solution

    We know that \( \cos(120^\circ) = -\frac{1}{2} \) and \( \cos(240^\circ) = -\frac{1}{2} \). Therefore, the solutions are \( x = 120^\circ \) and \( x = 240^\circ \).

  9. Solve \( 3\cos(x) - 2 = 0 \) for \( x \in [0^\circ, 360^\circ) \).
    Solution

    Isolate \( \cos(x) \): \( 3\cos(x) = 2 \), so \( \cos(x) = \frac{2}{3} \). Now, use the inverse cosine to find \( x = \cos^{-1}\left(\frac{2}{3}\right) \), which gives \( x \approx 48.19^\circ \). Since cosine is positive in the first and fourth quadrants, the second solution is \( x = 360^\circ - 48.19^\circ \approx 311.81^\circ \). Therefore, the solutions are \( x \approx 48.19^\circ \) and \( x \approx 311.81^\circ \).

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