Trigonometric Equations

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Answer Key & Explanations

Trigonometric Equations · Grade 11 · Worksheet 2

1. Charlotte is a sound engineer designing an audio cancellation system for a concert hall. Two speakers emit sound waves that interfere with each other. The combined sound pressure level at a certain seat is modeled by the equation 5sin²θ - 8sinθ + 3 = 0, where θ represents the phase shift between the two sound waves in radians, measured from 0 to 2π. Find all possible phase shifts θ in the interval [0, 2π) that result in zero sound pressure at that seat. Answer: θ = π/6, 5π/6, arcsin(3/5), π - arcsin(3/5) Solution: Start with the equation 5sin²θ - 8sinθ + 3 = 0. Factor the quadratic. We look for two numbers that multiply to 5*3 = 15 and add to -8.
   Full step-by-step solution

   Step 1: Start with the equation 5sin²θ - 8sinθ + 3 = 0. Step 2: Factor the quadratic. We look for two numbers that multiply to 5*3 = 15 and add to -8. These numbers are -5 and -3. Step 3: Rewrite: 5sin²θ - 5sinθ - 3sinθ + 3 = 0. Step 4: Factor by grouping: 5sinθ(sinθ - 1) - 3(sinθ - 1) = 0. Step 5: Factor out (sinθ - 1): (sinθ - 1)(5sinθ - 3) = 0. Step 6: Set each factor to zero: Case 1: sinθ - 1 = 0 → sinθ = 1. Case 2: 5sinθ - 3 = 0 → sinθ = 3/5. Step 7: Solve Case 1: sinθ = 1. In [0, 2π), the only solution is θ = π/2. Step 8: Solve Case 2: sinθ = 3/5. The reference angle is arcsin(3/5). Since sine is positive in quadrants I and II: θ = arcsin(3/5) and θ = π - arcsin(3/5). Step 9: All solutions in [0, 2π) are θ = π/2, arcsin(3/5), and π - arcsin(3/5). The answer is θ = π/2, arcsin(3/5), π - arcsin(3/5).

2. 2cos²(x) - √3cos(x) = 0 for x ∈ [0, 2π] Answer: π/6, π/2, 3π/2, 11π/6 Solution: Step 1: Factor the equation: 2cos²(x) - √3cos(x) = cos(x)(2cos(x) - √3) = 0 Step 2: Set each factor equal to zero: cos(x) = 0 OR 2cos(x) - √3 = 0 Step 3: Solve cos(x) = 0: x = π/2, 3π/2 Step 4: Solve 2cos(x) - √3 = 0: 2cos(x) = √3, cos(x) = √3/2, x = π/6, 11π/6 Step 5: Combine all solutions: x =…
   Full step-by-step solution

   Step 1: Factor the equation: 2cos²(x) - √3cos(x) = cos(x)(2cos(x) - √3) = 0 Step 2: Set each factor equal to zero: cos(x) = 0 OR 2cos(x) - √3 = 0 Step 3: Solve cos(x) = 0: x = π/2, 3π/2 Step 4: Solve 2cos(x) - √3 = 0: 2cos(x) = √3, cos(x) = √3/2, x = π/6, 11π/6 Step 5: Combine all solutions: x = π/6, π/2, 3π/2, 11π/6 Step 6: Verify all solutions are in [0, 2π]: All values are between 0 and 2π The solutions are π/6, π/2, 3π/2, and 11π/6.

3. 2sin²(x) - 1 = 0 for x ∈ [0, 2π] Answer: π/4, 3π/4, 5π/4, 7π/4 Solution: Step 1: Recognize that 2sin²(x) - 1 = 0 can be rewritten using the identity cos(2x) = 1 - 2sin²(x) Step 2: Rearrange the identity to get 2sin²(x) - 1 = -cos(2x) Step 3: Substitute into the equation: -cos(2x) = 0 Step 4: Multiply both sides by -1: cos(2x) = 0 Step 5: Solve for 2x: 2x = π/2, 3π/2,…
   Full step-by-step solution

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

4. 2cos²(x) - √3cos(x) = 0 for 0 ≤ x ≤ 2π Answer: π/2, 3π/2, π/6, 11π/6 Solution: Factor out cos(x) from the equation: cos(x)(2cos(x) - √3) = 0 Set each factor equal to zero: cos(x) = 0 OR 2cos(x) - √3 = 0 Solve cos(x) = 0: In the interval [0, 2π], cos(x) = 0 when x = π/2 and x = 3π/2 Solve 2cos(x) - √3 = 0: 2cos(x) = √3, so cos(x) = √3/2 In the interval [0, 2π], cos(x) =…
   Full step-by-step solution

   Step 1: Factor out cos(x) from the equation: cos(x)(2cos(x) - √3) = 0 Step 2: Set each factor equal to zero: cos(x) = 0 OR 2cos(x) - √3 = 0 Step 3: Solve cos(x) = 0: In the interval [0, 2π], cos(x) = 0 when x = π/2 and x = 3π/2 Step 4: Solve 2cos(x) - √3 = 0: 2cos(x) = √3, so cos(x) = √3/2 Step 5: In the interval [0, 2π], cos(x) = √3/2 when x = π/6 and x = 11π/6 Step 6: Combine all solutions: x = π/2, 3π/2, π/6, 11π/6 Final answer: π/2, 3π/2, π/6, 11π/6

5. Tane is an engineer designing a wave energy converter that harnesses power from ocean waves. The power output P(t) in kilowatts is modeled by the function P(t) = 12 sin(2t) + 9 cos(2t), where t is the time in seconds. To optimize energy storage, the system must activate when the power output is exactly 15 kilowatts. Find all times t in the interval [0, π) seconds when P(t) = 15. Answer: t = (1/2) * arcsin(3/5) and t = (1/2) * (π - arcsin(3/5)) Solution: Start with the equation 12 sin(2t) + 9 cos(2t) = 15. Rewrite 12 sin(2t) + 9 cos(2t) as a single sine function: R sin(2t + φ), where R = sqrt(12^2 + 9^2) = sqrt(144 + 81) = sqrt(225) = 15.
   Full step-by-step solution

   Step 1: Start with the equation 12 sin(2t) + 9 cos(2t) = 15. Step 2: Rewrite 12 sin(2t) + 9 cos(2t) as a single sine function: R sin(2t + φ), where R = sqrt(12^2 + 9^2) = sqrt(144 + 81) = sqrt(225) = 15. Step 3: Find φ such that sin φ = 9/15 = 3/5 and cos φ = 12/15 = 4/5. So φ = arcsin(3/5). Step 4: The equation becomes 15 sin(2t + φ) = 15, so sin(2t + φ) = 1. Step 5: sin(θ) = 1 when θ = π/2 + 2πk, where k is an integer. Thus 2t + φ = π/2 + 2πk. Step 6: Solve for t: 2t = π/2 - φ + 2πk, so t = π/4 - φ/2 + πk. Step 7: Since φ = arcsin(3/5), t = π/4 - (1/2)arcsin(3/5) + πk. Step 8: For k = 0: t = π/4 - (1/2)arcsin(3/5). For k = 1: t = π/4 - (1/2)arcsin(3/5) + π = 5π/4 - (1/2)arcsin(3/5). Step 9: Check the interval [0, π). For k = 0: t is in [0, π). For k = 1: t is greater than π. So only k = 0 gives a solution. Step 10: Alternatively, using the identity sin(2t + φ) = 1 gives 2t + φ = π/2, so 2t = π/2 - φ, t = π/4 - φ/2. With φ = arcsin(3/5), t = (1/2)(π/2 - arcsin(3/5)). The solutions in [0, π) are t = (1/2)(π/2 - arcsin(3/5)) and t = (1/2)(π - arcsin(3/5)).

6. 2sin²(x) - sin(x) - 1 = 0 for 0 ≤ x < 2π Answer: x = π/2, 7π/6, 11π/6 Solution: 2 sin²(x) - sin(x) - 1 = 0, for 0 ≤ x < 2π. Let u = sin(x). 2u² - u - 1 = 0.
   Full step-by-step solution

   Let's solve the equation step by step. We are given: 2 sin²(x) - sin(x) - 1 = 0, for 0 ≤ x < 2π. --- **Step 1: Substitute to simplify** Let u = sin(x). Then the equation becomes: 2u² - u - 1 = 0. --- **Step 2: Solve the quadratic equation** We can factor: 2u² - u - 1 = 0 (2u + 1)(u - 1) = 0. So: 2u + 1 = 0 → u = -1/2 or u - 1 = 0 → u = 1. --- **Step 3: Back substitute u = sin(x)** Case 1: sin(x) = 1 In 0 ≤ x < 2π, sin(x) = 1 only at x = π/2. Case 2: sin(x) = -1/2 We know sin(x) is negative in Quadrants III and IV. Reference angle: sin(π/6) = 1/2, so for sin(x) = -1/2: x = π + π/6 = 7π/6, x = 2π - π/6 = 11π/6. --- **Step 4: List all solutions in the interval** x = π/2, 7π/6, 11π/6. --- **Final answer:** x = π/2, 7π/6, 11π/6

7. Matiu is an acoustic engineer designing a noise-canceling system for a factory floor. Two machines produce sound waves that interfere, and the combined sound pressure at a sensor is modeled by the equation 2sin²θ - 3sinθ + 1 = 0, where θ is the phase difference in radians between the waves, measured from 0 to 2π. Find all possible phase differences θ in the interval [0, 2π) that result in zero sound pressure at the sensor. Answer: θ = π/6, 5π/6, π/2 Solution: Start with the equation 2sin²θ - 3sinθ + 1 = 0. Factor the quadratic in sinθ. We look for two numbers that multiply to 2 * 1 = 2 and add to -3.
   Full step-by-step solution

   Step 1: Start with the equation 2sin²θ - 3sinθ + 1 = 0. Step 2: Factor the quadratic in sinθ. We look for two numbers that multiply to 2 * 1 = 2 and add to -3. These numbers are -2 and -1. Step 3: Rewrite: 2sin²θ - 2sinθ - sinθ + 1 = 0. Step 4: Factor by grouping: 2sinθ(sinθ - 1) - 1(sinθ - 1) = 0. Step 5: Factor out (sinθ - 1): (sinθ - 1)(2sinθ - 1) = 0. Step 6: Set each factor to zero: Case 1: sinθ - 1 = 0 → sinθ = 1. Case 2: 2sinθ - 1 = 0 → sinθ = 1/2. Step 7: Solve Case 1: sinθ = 1. In [0, 2π), the only solution is θ = π/2. Step 8: Solve Case 2: sinθ = 1/2. Since sine is positive in quadrants I and II, the reference angle is π/6. So: θ = π/6 and θ = π - π/6 = 5π/6. Step 9: All solutions in [0, 2π) are θ = π/6, 5π/6, π/2. The answer is θ = π/6, 5π/6, π/2.