Trigonometric Equations

Grade 11 · algebra · 100 practice problems · read aloud

🔊 Listen to this explanation

What Are Trigonometric Equations? 🤔

A trigonometric equation is an equation that involves trigonometric functions (like sine, cosine, and tangent) of a variable. Solving them means finding all angles that make the equation true. This is crucial for modeling real-world periodic phenomena like sound waves, light waves, and seasonal changes.

How to Solve Trigonometric Equations: A Step-by-Step Guide

  1. Isolate the trigonometric function (e.g., sin θ, cos θ).
  2. Use the inverse trigonometric function to find the principal angle(s).
  3. Determine the quadrants where the function has the correct sign (positive or negative).
  4. Find all solutions within one period (0 to 2π). Use reference angles.
  5. Generalize the solution by adding + 2πk (or + πk for tangent) to account for all coterminal angles, where k is any integer.

Worked Examples

Example 1: Basic Equation

Solve: 2 sin θ - 1 = 0 for 0 ≤ θ < 2π

  1. Isolate: 2 sin θ = 1 → sin θ = 1/2
  2. Principal Angle: θ = sin⁻¹(1/2) = π/6
  3. Quadrants: Sine is positive in QI and QII.
  4. Find all solutions: QI: θ = π/6, QII: θ = π - π/6 = 5π/6
  5. Solution Set: {π/6, 5π/6}

Example 2: Equation with a Multiple Angle

Solve: cos(2x) = -√3/2 for 0 ≤ x < 2π

  1. Isolate: cos(2x) = -√3/2
  2. Substitute: Let u = 2x. Solve cos u = -√3/2.
  3. Principal Angle: u = cos⁻¹(√3/2) = π/6. Cosine is negative in QII and QIII.
  4. Find u: u = π - π/6 = 5π/6 and u = π + π/6 = 7π/6.
  5. Generalize u: u = 5π/6 + 2πk and u = 7π/6 + 2πk
  6. Solve for x: Since u = 2x, x = u/2. So, x = 5π/12 + πk and x = 7π/12 + πk.
  7. Solutions in [0, 2π): For k=0: x=5π/12, 7π/12. For k=1: x=17π/12, 19π/12.

Common Mistakes to Avoid 🚫

  • Forgetting the second quadrant: When you take an inverse sine or cosine, your calculator gives you one answer. Always remember to find the second solution in the other relevant quadrant.
  • Misapplying the period: The period of sin and cos is 2π, but the period of tan is only π. Adding 2πk to a tangent solution will make you miss half the answers!
  • Ignoring the domain: Always check if the problem asks for solutions in a specific interval (like [0, 2π)) or the general solution.

Tips & Tricks

  • ASTC Rule: Remember "All Students Take Calculus" to recall which trig functions are positive in each quadrant (QI: All, QII: Sine, QIII: Tangent, QIV: Cosine).
  • Unit Circle is Key: Knowing your unit circle values for 0, π/6, π/4, π/3, and π/2 is essential for solving these equations quickly and accurately.
  • Check Your Answers: Plug your solutions back into the original equation to verify they work!

How to Practice

To master trigonometric equations, practice is essential. Start with simple equations that isolate one trig function. Then, move on to equations that require factoring (e.g., sin² x - sin x = 0) and those with multiple angles. Always practice finding both specific solutions within an interval and the general solution. Use online resources and your textbook for a variety of problems.

Practice problems

6 of the 100, worked through step by step — try them before opening the answer.

1 sin(2x) = cos(x) for 0 ≤ x ≤ 2π

Hint: Use a double-angle identity to rewrite the equation in terms of a single trigonometric function, then solve the resulting equation.

Show the answer

Answer: π/6, π/2, 5π/6, 3π/2

  1. Use the double-angle identity: sin(2x) = 2sin(x)cos(x)
  2. Substitute into the equation: 2sin(x)cos(x) = cos(x)
  3. Rearrange: 2sin(x)cos(x) - cos(x) = 0
  4. Factor: cos(x)(2sin(x) - 1) = 0
  5. Set each factor equal to zero: cos(x) = 0 OR 2sin(x) - 1 = 0
  6. Solve cos(x) = 0 in [0, 2π]: x = π/2, 3π/2
  7. Solve 2sin(x) - 1 = 0: sin(x) = 1/2
  8. Solve sin(x) = 1/2 in [0, 2π]: x = π/6, 5π/6
  9. Combine all solutions: x = π/6, π/2, 5π/6, 3π/2

2 2sin²(x) - 1 = 0 for x ∈ [0, 2π]

Hint: This equation can be rewritten using a double-angle identity. Consider how cosine of 2x relates to sine squared of x.

Show the answer

Answer: π/4, 3π/4, 5π/4, 7π/4

  1. Recognize that 2sin²(x) - 1 = 0 can be rewritten using the identity cos(2x) = 1 - 2sin²(x)
  2. Rearrange the identity to get 2sin²(x) - 1 = -cos(2x)
  3. Substitute into the equation: -cos(2x) = 0
  4. Multiply both sides by -1: cos(2x) = 0
  5. Solve for 2x: 2x = π/2, 3π/2, 5π/2, 7π/2
  6. Divide by 2: x = π/4, 3π/4, 5π/4, 7π/4
  7. Verify all solutions are in [0, 2π] The solutions are x = π/4, 3π/4, 5π/4, 7π/4.

3 sin(2x) = cos(x) for x ∈ [0, 2π]

Hint: Use a double-angle identity to rewrite the equation in terms of a single trigonometric function. Consider all possible cases when a product equals zero.

Show the answer

Answer: π/6, π/2, 5π/6, 3π/2

  1. Use the double-angle identity: sin(2x) = 2sin(x)cos(x)
  2. Substitute into the equation: 2sin(x)cos(x) = cos(x)
  3. Rearrange: 2sin(x)cos(x) - cos(x) = 0
  4. Factor: cos(x)(2sin(x) - 1) = 0
  5. Set each factor equal to zero: Case 1: cos(x) = 0 Case 2: 2sin(x) - 1 = 0
  6. Solve Case 1: cos(x) = 0 in [0, 2π] gives x = π/2, 3π/2
  7. Solve Case 2: 2sin(x) - 1 = 0 → sin(x) = 1/2 in [0, 2π] gives x = π/6, 5π/6
  8. Combine all solutions: x = π/6, π/2, 5π/6, 3π/2

4 3tan²(x) - 1 = 0 for x ∈ [0, 2π]

Hint: Isolate the tangent squared term first, then take the square root. Remember that the tangent function is positive in quadrants I and III, and negative in quadrants II and IV.

Show the answer

Answer: x = π/6, 5π/6, 7π/6, 11π/6

  1. Add 1 to both sides: 3tan²(x) = 1
  2. Divide by 3: tan²(x) = 1/3
  3. Take the square root of both sides: tan(x) = ±1/√3
  4. Rationalize: tan(x) = ±√3/3
  5. The reference angle where tan(θ) = √3/3 is θ = π/6.
  6. For tan(x) = √3/3 (positive), solutions are in quadrants I and III: Quadrant I: x = π/6 Quadrant III: x = π + π/6 = 7π/6
  7. For tan(x) = -√3/3 (negative), solutions are in quadrants II and IV: Quadrant II: x = π - π/6 = 5π/6 Quadrant IV: x = 2π - π/6 = 11π/6
  8. All solutions in [0, 2π] are x = π/6, 5π/6, 7π/6, 11π/6.

The answer is x = π/6, 5π/6, 7π/6, 11π/6.

5 sin(2x) - cos(x) = 0 for 0 ≤ x ≤ 2π

Hint: Use a trigonometric identity to rewrite sin(2x) in terms of sin(x) and cos(x), then factor the resulting equation.

Show the answer

Answer: π/6, π/2, 5π/6, 3π/2

  1. Use the double-angle identity: sin(2x) = 2sin(x)cos(x) The equation becomes: 2sin(x)cos(x) - cos(x) = 0
  2. Factor out cos(x): cos(x)(2sin(x) - 1) = 0
  3. Set each factor equal to zero: Case 1: cos(x) = 0 In the interval [0, 2π], cos(x) = 0 when x = π/2 and x = 3π/2 Case 2: 2sin(x) - 1 = 0 2sin(x) = 1 sin(x) = 1/2 In the interval [0, 2π], sin(x) = 1/2 when x = π/6 and x = 5π/6
  4. Combine all solutions: x = π/6, π/2, 5π/6, 3π/2

The answer is π/6, π/2, 5π/6, 3π/2.

6 sin(2x) + cos(x) = 0 for x ∈ [0, 2π]

Hint: Use a trigonometric identity to rewrite sin(2x) in terms of sin(x) and cos(x), then factor the resulting equation.

Show the answer

Answer: π/2, 3π/2, 7π/6, 11π/6

  1. Use the double-angle identity: sin(2x) = 2sin(x)cos(x)
  2. Substitute into the equation: 2sin(x)cos(x) + cos(x) = 0
  3. Factor out cos(x): cos(x)(2sin(x) + 1) = 0
  4. Set each factor equal to zero: Case 1: cos(x) = 0 → x = π/2, 3π/2 Case 2: 2sin(x) + 1 = 0 → sin(x) = -1/2 → x = 7π/6, 11π/6
  5. All solutions in [0, 2π] are: π/2, 3π/2, 7π/6, 11π/6
Practise this topic — 10 free problems, no signup →