Solving Trigonometric Equations Find Exact Solutions On [0, 2π)

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In the realm of trigonometry, determining the exact solutions of equations within a specified interval is a fundamental skill. This article delves into the process of finding all exact solutions for the equation 2sin(θ) = -√3 on the interval [0, 2π). We'll explore the underlying concepts, step-by-step methodology, and provide a comprehensive understanding of how to tackle such problems.

Understanding Trigonometric Equations and Solutions

Trigonometric equations involve trigonometric functions such as sine, cosine, tangent, and their reciprocals. Solving these equations entails finding the angles (θ) that satisfy the given equation. The interval [0, 2π) represents one complete revolution around the unit circle, encompassing all possible angles from 0 to 360 degrees (excluding 360 degrees itself). Finding solutions within this interval ensures we capture all unique solutions within a single cycle.

When we solve trigonometric equations, we're essentially looking for the angles on the unit circle where the trigonometric function's value matches the given value. For instance, in our case, we're seeking angles where the sine function equals -√3/2. The unit circle serves as a visual aid, helping us identify these angles based on their coordinates. Remember that the sine function corresponds to the y-coordinate on the unit circle.

Step-by-Step Solution

Let's embark on a step-by-step journey to solve the equation 2sin(θ) = -√3 on the interval [0, 2π). This methodical approach will not only lead us to the correct solutions but also reinforce the underlying principles of solving trigonometric equations.

1. Isolate the Trigonometric Function

The initial step involves isolating the trigonometric function on one side of the equation. In our case, we have 2sin(θ) = -√3. To isolate sin(θ), we divide both sides of the equation by 2:

sin(θ) = -√3 / 2

This step simplifies the equation and brings us closer to identifying the angles that satisfy the condition.

2. Identify Reference Angle

The reference angle is the acute angle formed between the terminal side of the angle and the x-axis. To find the reference angle, we consider the absolute value of the trigonometric function's value. In our case, we have sin(θ) = -√3 / 2. The absolute value is √3 / 2. We need to determine the angle whose sine is √3 / 2. Recall the special right triangles, particularly the 30-60-90 triangle. The sine of 60 degrees (π/3 radians) is √3 / 2. Therefore, the reference angle is π/3.

The reference angle acts as a building block for finding all solutions within the interval. It allows us to pinpoint the angles in different quadrants that have the same trigonometric function value (in absolute terms).

3. Determine Quadrants

The next crucial step is to identify the quadrants where the sine function is negative. Recall the acronym ASTC, which stands for "All Students Take Calculus." This helps us remember which trigonometric functions are positive in each quadrant:

  • Quadrant I (All): All trigonometric functions are positive.
  • Quadrant II (Sine): Sine is positive.
  • Quadrant III (Tangent): Tangent is positive.
  • Quadrant IV (Cosine): Cosine is positive.

Since we have sin(θ) = -√3 / 2, and sine is negative, the solutions will lie in Quadrants III and IV.

4. Find Solutions in the Identified Quadrants

Now, we utilize the reference angle and the quadrants to find the solutions within the interval [0, 2π). In each quadrant, we'll construct the angle using the reference angle.

  • Quadrant III: In Quadrant III, the angle is given by π + reference angle. Therefore,

    θ = π + π/3 = 4π/3

  • Quadrant IV: In Quadrant IV, the angle is given by 2π - reference angle. Therefore,

    θ = 2π - π/3 = 5π/3

These two angles, 4π/3 and 5π/3, are the solutions to the equation 2sin(θ) = -√3 within the interval [0, 2π).

5. Verify the Solutions

It's always a good practice to verify the solutions by substituting them back into the original equation. Let's check:

  • For θ = 4π/3:

    2sin(4π/3) = 2(-√3/2) = -√3 (Correct)

  • For θ = 5π/3:

    2sin(5π/3) = 2(-√3/2) = -√3 (Correct)

Both solutions satisfy the original equation.

General Solutions vs. Solutions in [0, 2π)

It's important to distinguish between general solutions and solutions within a specific interval. General solutions represent all possible angles that satisfy the equation, typically expressed using the form θ + 2πk, where k is an integer. This accounts for the periodic nature of trigonometric functions. However, when we're asked for solutions in [0, 2π), we're only interested in the solutions within one complete cycle.

In our case, the general solutions would be 4π/3 + 2πk and 5π/3 + 2πk. But the solutions we found (4π/3 and 5π/3) are the specific solutions that fall within the interval [0, 2π).

Common Mistakes to Avoid

Solving trigonometric equations can be tricky, and several common mistakes can lead to incorrect answers. Being aware of these pitfalls can help you avoid them:

  • Forgetting the ± Sign: When taking the square root, remember to consider both positive and negative roots. For example, if sin²(θ) = 1/4, then sin(θ) = ±1/2.
  • Incorrectly Identifying Quadrants: Make sure you correctly identify the quadrants where the trigonometric function has the given sign.
  • Using Degrees Instead of Radians (or Vice Versa): Ensure you're using the correct unit of measurement for angles. If the interval is given in terms of π (radians), your solutions should also be in radians.
  • Missing Solutions: Double-check that you've found all solutions within the specified interval. There might be multiple solutions within one cycle.
  • Not Verifying Solutions: Always verify your solutions by substituting them back into the original equation.

Practice Problems

To solidify your understanding, let's explore some practice problems similar to the one we solved. Working through these examples will enhance your skills and confidence in solving trigonometric equations.

Practice Problem 1

Solve the equation cos(θ) = -1/2 on the interval [0, 2π).

Solution:

  1. The cosine function is already isolated.
  2. The reference angle is π/3 (since cos(π/3) = 1/2).
  3. Cosine is negative in Quadrants II and III.
  4. Solutions:
    • Quadrant II: θ = π - π/3 = 2π/3
    • Quadrant III: θ = π + π/3 = 4π/3
  5. The solutions are 2π/3 and 4π/3.

Practice Problem 2

Solve the equation 2sin(θ) + 1 = 0 on the interval [0, 2π).

Solution:

  1. Isolate sin(θ): 2sin(θ) = -1 => sin(θ) = -1/2
  2. The reference angle is π/6 (since sin(π/6) = 1/2).
  3. Sine is negative in Quadrants III and IV.
  4. Solutions:
    • Quadrant III: θ = π + π/6 = 7π/6
    • Quadrant IV: θ = 2π - π/6 = 11π/6
  5. The solutions are 7π/6 and 11π/6.

Practice Problem 3

Solve the equation √3 tan(θ) = 1 on the interval [0, 2π).

Solution:

  1. Isolate tan(θ): tan(θ) = 1/√3 = √3/3
  2. The reference angle is π/6 (since tan(π/6) = √3/3).
  3. Tangent is positive in Quadrants I and III.
  4. Solutions:
    • Quadrant I: θ = π/6
    • Quadrant III: θ = π + π/6 = 7π/6
  5. The solutions are π/6 and 7π/6.

Conclusion

Mastering the art of finding exact solutions for trigonometric equations on the interval [0, 2π) requires a firm grasp of the unit circle, reference angles, and quadrant rules. By following a systematic approach, you can confidently tackle a wide range of trigonometric equations. Remember to practice regularly and verify your solutions to ensure accuracy. The journey through trigonometry can be challenging, but with consistent effort and a clear understanding of the fundamental concepts, you can unlock its beauty and power.

By understanding the concepts of reference angles, quadrants, and the unit circle, we can confidently solve trigonometric equations and find exact solutions within specified intervals. This skill is crucial for various applications in mathematics, physics, and engineering. Remember to practice regularly and verify your solutions to solidify your understanding.