Solve The Equation On The Interval 0 2pi

Solving Trigonometric Equations on the Interval [0, 2π)

Solving trigonometric equations is a fundamental skill in mathematics, particularly in calculus and physics. This comprehensive guide will walk you through various techniques for solving trigonometric equations, focusing specifically on solutions within the interval [0, 2π). We'll explore different approaches, including using trigonometric identities, factoring, and the quadratic formula, providing numerous examples to solidify your understanding.

Understanding the Unit Circle and Trigonometric Functions

Before diving into solving equations, let's refresh our understanding of the unit circle and the six trigonometric functions: sine (sin), cosine (cos), tangent (tan), cosecant (csc), secant (sec), and cotangent (cot). The unit circle is a circle with a radius of 1, centered at the origin (0, 0) of the coordinate plane. Each point on the unit circle can be represented by its coordinates (cos θ, sin θ), where θ is the angle formed between the positive x-axis and the line segment connecting the origin to the point.

Key Trigonometric Identities

Several identities are crucial for simplifying and solving trigonometric equations:

• Pythagorean Identities:

  • sin²θ + cos²θ = 1
  • 1 + tan²θ = sec²θ
  • 1 + cot²θ = csc²θ

• Sum and Difference Formulas:

  • sin(A ± B) = sin A cos B ± cos A sin B
  • cos(A ± B) = cos A cos B ∓ sin A sin B
  • tan(A ± B) = (tan A ± tan B) / (1 ∓ tan A tan B)

• Double Angle Formulas:

  • sin 2θ = 2 sin θ cos θ
  • cos 2θ = cos²θ - sin²θ = 2cos²θ - 1 = 1 - 2sin²θ
  • tan 2θ = 2 tan θ / (1 - tan²θ)

• Half Angle Formulas:

  • sin (θ/2) = ±√[(1 - cos θ)/2]
  • cos (θ/2) = ±√[(1 + cos θ)/2]
  • tan (θ/2) = ±√[(1 - cos θ)/(1 + cos θ)] = sin θ / (1 + cos θ) = (1 - cos θ) / sin θ

Mastering these identities is essential for effectively manipulating trigonometric equations.

Solving Basic Trigonometric Equations

Let's begin with straightforward examples:

Example 1: sin θ = 1/2

To solve this, we consider the unit circle. The sine function represents the y-coordinate. Where on the unit circle is the y-coordinate equal to 1/2? This occurs at θ = π/6 and θ = 5π/6. Therefore, the solutions on the interval [0, 2π) are π/6 and 5π/6.

Example 2: cos θ = -√3/2

Similarly, the cosine function represents the x-coordinate. On the unit circle, the x-coordinate is -√3/2 at θ = 5π/6 and θ = 7π/6. Thus, the solutions in the interval [0, 2π) are 5π/6 and 7π/6.

Example 3: tan θ = 1

The tangent function is the ratio of sine to cosine (sin θ / cos θ). The tangent is positive in the first and third quadrants. On the unit circle, tan θ = 1 at θ = π/4 and θ = 5π/4. The solutions on the interval [0, 2π) are therefore π/4 and 5π/4.

Solving More Complex Trigonometric Equations

More complex equations often require the use of trigonometric identities and algebraic manipulation.

Example 4: 2sin²θ - sin θ - 1 = 0

This equation is a quadratic equation in terms of sin θ. We can factor it as:

(2sin θ + 1)(sin θ - 1) = 0

This gives us two possible equations:

2sin θ + 1 = 0 => sin θ = -1/2 sin θ - 1 = 0 => sin θ = 1

Solving for sin θ = -1/2, we get θ = 7π/6 and θ = 11π/6. Solving for sin θ = 1, we get θ = π/2.

Therefore, the solutions on the interval [0, 2π) are π/2, 7π/6, and 11π/6.

Example 5: cos²θ + cos θ - 2 = 0

This is again a quadratic equation, this time in cos θ. Factoring, we have:

(cos θ + 2)(cos θ - 1) = 0

This leads to:

cos θ + 2 = 0 => cos θ = -2 (This has no solution since -1 ≤ cos θ ≤ 1) cos θ - 1 = 0 => cos θ = 1

Therefore, the only solution in the interval [0, 2π) is θ = 0.

Example 6: sin 2θ = cos θ

Using the double angle formula for sin 2θ (sin 2θ = 2 sin θ cos θ), we rewrite the equation as:

2 sin θ cos θ = cos θ

Subtracting cos θ from both sides:

2 sin θ cos θ - cos θ = 0

Factoring out cos θ:

cos θ (2 sin θ - 1) = 0

This gives two equations:

cos θ = 0 => θ = π/2, 3π/2 2 sin θ - 1 = 0 => sin θ = 1/2 => θ = π/6, 5π/6

Therefore, the solutions on the interval [0, 2π) are π/6, π/2, 5π/6, and 3π/2.

Solving Equations Using the Quadratic Formula

For more complex quadratic equations that are not easily factored, the quadratic formula can be applied. Remember that the quadratic formula solves equations of the form ax² + bx + c = 0, where:

x = [-b ± √(b² - 4ac)] / 2a

Example 7: 3cos²θ - 5cos θ + 2 = 0

This is a quadratic equation in cos θ. Using the quadratic formula (a=3, b=-5, c=2):

cos θ = [5 ± √((-5)² - 4 * 3 * 2)] / (2 * 3) cos θ = [5 ± √(25 - 24)] / 6 cos θ = [5 ± 1] / 6

This gives two solutions:

cos θ = 1 => θ = 0 cos θ = 2/3

To find the solutions for cos θ = 2/3, we use the inverse cosine function (arccos):

θ = arccos(2/3) ≈ 0.841 radians and θ = 2π - 0.841 ≈ 5.442 radians

Therefore, the approximate solutions on the interval [0, 2π) are 0, 0.841, and 5.442.

Equations Involving Multiple Angles

Equations with multiple angles (like 3θ, 5θ, etc.) require an extra step. Solve for the multiple angle first, then divide to find θ. Remember to consider all possible solutions within the extended interval.

Example 8: sin 3θ = 1/2

First, solve for 3θ:

3θ = π/6 + 2kπ or 3θ = 5π/6 + 2kπ, where k is an integer.

Now, divide by 3:

θ = π/18 + (2kπ)/3 or θ = 5π/18 + (2kπ)/3

To find solutions in the interval [0, 2π), we substitute different integer values of k:

For k = 0: θ = π/18, 5π/18 For k = 1: θ = π/18 + 2π/3 = 13π/18, 5π/18 + 2π/3 = 17π/18 For k = 2: θ = π/18 + 4π/3 = 25π/18, 5π/18 + 4π/3 = 29π/18 (These are greater than 2π)

Therefore, the solutions on the interval [0, 2π) are π/18, 5π/18, 13π/18, and 17π/18.

Conclusion

Solving trigonometric equations on the interval [0, 2π) involves a combination of understanding trigonometric identities, algebraic manipulation, and utilizing the unit circle. By mastering these techniques and practicing with various examples, you can confidently tackle a wide range of trigonometric equation problems. Remember to always check your solutions by substituting them back into the original equation and verifying that they satisfy the equation within the specified interval. Consistent practice and a solid grasp of the fundamental concepts are key to success in this area of mathematics.