Trigonometric Equations

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Trigonometric Equations · Grade 11 · Worksheet 2

1. Aroha is an aerospace engineer modeling the vibration of a satellite's solar panel. The panel's angular displacement θ (in radians, from 0 to 2π) is governed by the equation 9sin²θ - 12sinθ + 4 = 0. Find all possible angular displacements θ in the interval [0, 2π) that satisfy this equation. Answer: θ = π/6, 5π/6 Solution: The equation is 9sin²θ - 12sinθ + 4 = 0. Recognize this as a quadratic in sinθ. Check if it factors as a perfect square: (3sinθ - 2)² = 9sin²θ - 12sinθ + 4.
   Full step-by-step solution

   Step 1: The equation is 9sin²θ - 12sinθ + 4 = 0. Step 2: Recognize this as a quadratic in sinθ. Check if it factors as a perfect square: (3sinθ - 2)² = 9sin²θ - 12sinθ + 4. Yes, it matches. Step 3: So (3sinθ - 2)² = 0. Step 4: Take square root: 3sinθ - 2 = 0. Step 5: Solve: 3sinθ = 2, so sinθ = 2/3. Step 6: The reference angle is arcsin(2/3). Since 2/3 is positive, sine is positive in quadrants I and II. Step 7: In quadrant I: θ = arcsin(2/3). Step 8: In quadrant II: θ = π - arcsin(2/3). Step 9: Both solutions are in [0, 2π). The final answer is θ = arcsin(2/3) and θ = π - arcsin(2/3).

2. An engineer is designing a suspension bridge where the main cable forms a parabolic shape approximated by the function y = 25 cos(πx/100) + 15, where y is the height in meters above the road deck and x is the horizontal distance in meters from the center of the bridge. The engineer needs to install vertical support cables at the points where the main cable is exactly 30 meters above the road deck. Find all values of x between -50 and 50 meters where this occurs. Answer: ±33.33 meters Solution: Trigonometric equations often model periodic phenomena like waves, vibrations, or in this case, bridge cables.
   Full step-by-step solution

   Trigonometric equations often model periodic phenomena like waves, vibrations, or in this case, bridge cables. To solve them, we use algebraic manipulation to isolate the trigonometric term, then apply inverse trigonometric functions. Since these functions are periodic, we must consider all possible solutions within the given domain, often using knowledge of the unit circle and reference angles. The solutions represent the input values where the function reaches a specific output value.

3. 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.

4. Olivia is an audio engineer calibrating a parametric equalizer for a recording studio. The equalizer's frequency response is modeled by the equation 2cos²θ + 5sinθ - 4 = 0, where θ represents the phase shift in radians applied to a test tone, measured from 0 to 2π. Find all phase shifts θ in the interval [0, 2π) that make the frequency response zero. Answer: θ = π/6, 5π/6 Solution: Start with the equation 2cos²θ + 5sinθ - 4 = 0. Use the identity cos²θ = 1 - sin²θ to rewrite: 2(1 - sin²θ) + 5sinθ - 4 = 0. Distribute: 2 - 2sin²θ + 5sinθ - 4 = 0.
   Full step-by-step solution

   Step 1: Start with the equation 2cos²θ + 5sinθ - 4 = 0. Step 2: Use the identity cos²θ = 1 - sin²θ to rewrite: 2(1 - sin²θ) + 5sinθ - 4 = 0. Step 3: Distribute: 2 - 2sin²θ + 5sinθ - 4 = 0. Step 4: Simplify: -2sin²θ + 5sinθ - 2 = 0. Step 5: Multiply both sides by -1: 2sin²θ - 5sinθ + 2 = 0. Step 6: Factor the quadratic: (2sinθ - 1)(sinθ - 2) = 0. Step 7: Set each factor to zero: Case 1: 2sinθ - 1 = 0 → sinθ = 1/2. Case 2: sinθ - 2 = 0 → sinθ = 2. Step 8: Since sinθ = 2 has no solution (sine is always between -1 and 1), we discard it. Step 9: Solve sinθ = 1/2 in [0, 2π). The reference angle is π/6. Sine is positive in quadrants I and II. Quadrant I: θ = π/6. Quadrant II: θ = π - π/6 = 5π/6. Step 10: The solutions are θ = π/6 and θ = 5π/6. The answer is θ = π/6, 5π/6.

5. An architect is designing a suspension bridge where the main cable forms a parabolic shape. The cable's height above the roadway can be modeled by the function h(x) = 50 - 0.02x², where x is the horizontal distance from the center of the bridge in meters. The supporting cables are vertical and spaced every 10 meters. At what horizontal distances from the center will a supporting cable be exactly 32 meters long? Answer: 30 meters and -30 meters Solution: h(x) = 50 - 0.02x² Here, h(x) is the height of the main cable above the roadway at a horizontal distance x from the center.
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Understand the problem** The parabolic cable height is given by h(x) = 50 - 0.02x² Here, h(x) is the height of the main cable above the roadway at a horizontal distance x from the center. The supporting cables are vertical, so their length is simply h(x) (since roadway is at height 0). We are told: supporting cable length = 32 m. So h(x) = 32. --- **Step 2: Set up the equation** h(x) = 50 - 0.02x² = 32 --- **Step 3: Solve for x** 50 - 0.02x² = 32 Subtract 50 from both sides: -0.02x² = 32 - 50 -0.02x² = -18 --- **Step 4: Divide by -0.02** x² = (-18) / (-0.02) x² = 18 / 0.02 x² = 1800 / 2 x² = 900 --- **Step 5: Take square root** x = ±√900 x = ±30 --- **Step 6: Interpret the result** The supporting cables at x = 30 m and x = -30 m from the center will be 32 m long. --- **Final answer:** 30 meters and -30 meters

6. sin(2x) + cos(x) = 0 for x ∈ [0, 2π] Answer: π/2, 3π/2, 7π/6, 11π/6 Solution: Use the double-angle identity: sin(2x) = 2sin(x)cos(x) Substitute into the equation: 2sin(x)cos(x) + cos(x) = 0 Factor out cos(x): cos(x)(2sin(x) + 1) = 0 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 All solutions in [0, 2π] are: π/2, 3π/2, 7π/6, 11π/6
   Full step-by-step solution

   Step 1: Use the double-angle identity: sin(2x) = 2sin(x)cos(x) Step 2: Substitute into the equation: 2sin(x)cos(x) + cos(x) = 0 Step 3: Factor out cos(x): cos(x)(2sin(x) + 1) = 0 Step 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 Step 5: All solutions in [0, 2π] are: π/2, 3π/2, 7π/6, 11π/6