Trigonometric Equations

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Answer Key & Explanations

Trigonometric Equations · Grade 11 · Worksheet 1

1. Ava is an astronomer studying the light curve of a variable star. The brightness of the star over time is modeled by the equation B(t) = 6 sin^2(t) - 11 sin(t) + 6, where t is the phase angle in radians measured from 0 to 2π. Ava needs to find all phase angles t in the interval [0, 2π) where the brightness drops to zero. Solve the trigonometric equation to help Ava identify these critical phase angles. Answer: t = π/6, 5π/6, π/2 Solution: The equation is 6 sin^2(t) - 11 sin(t) + 6 = 0. Factor the quadratic in terms of sin(t). Multiply 6 * 6 = 36.
   Full step-by-step solution

   Step 1: The equation is 6 sin^2(t) - 11 sin(t) + 6 = 0. Step 2: Factor the quadratic in terms of sin(t). Multiply 6 * 6 = 36. Find two numbers that multiply to 36 and add to -11: these are -9 and -2. Step 3: Rewrite the middle term: 6 sin^2(t) - 9 sin(t) - 2 sin(t) + 6 = 0. Step 4: Factor by grouping: 3 sin(t)(2 sin(t) - 3) - 1(2 sin(t) - 3) = 0. Step 5: Factor out (2 sin(t) - 3): (2 sin(t) - 3)(3 sin(t) - 1) = 0. Step 6: Set each factor to zero. Case 1: 2 sin(t) - 3 = 0 → sin(t) = 3/2. Case 2: 3 sin(t) - 1 = 0 → sin(t) = 1/3. Step 7: Check Case 1: sin(t) = 3/2 is impossible because sin(t) is always between -1 and 1. So no solutions from this case. Step 8: Solve Case 2: sin(t) = 1/3. The reference angle is arcsin(1/3). Since sine is positive in quadrants I and II, the solutions in [0, 2π) are: t = arcsin(1/3) and t = π - arcsin(1/3). Step 9: These are the exact solutions: t = arcsin(1/3) and t = π - arcsin(1/3). The answer is t = arcsin(1/3), π - arcsin(1/3).

2. A radio tower is anchored by two guy wires on opposite sides. The first wire makes a 60° angle with the ground and is attached 50 meters from the tower's base. The second wire makes a 45° angle with the ground. If both wires reach the same point at the top of the tower, what is the height of the tower? Answer: 50√3 Solution: We have a radio tower of height \( h \) (unknown). Two guy wires on opposite sides are attached to the top of the tower and anchored to the ground. - First wire: angle with ground = 60°, anchor point is 50 m from base of tower.
   Full step-by-step solution

   Let's solve step by step. --- **Step 1: Understand the problem** We have a radio tower of height \( h \) (unknown). Two guy wires on opposite sides are attached to the top of the tower and anchored to the ground. - First wire: angle with ground = 60°, anchor point is 50 m from base of tower. - Second wire: angle with ground = 45°, anchor point is some distance \( d \) from base. - Both wires reach the same point at the top. --- **Step 2: Set up equations for height** For the first wire: Angle with ground = 60°, horizontal distance = 50 m. Using tangent: \(\tan 60^\circ = \frac{h}{50}\) \(\tan 60^\circ = \sqrt{3}\) So: \( h = 50 \times \sqrt{3} \) --- **Step 3: Check second wire** For the second wire: Angle with ground = 45°, horizontal distance = \( d \). Using tangent: \(\tan 45^\circ = \frac{h}{d}\) \(\tan 45^\circ = 1\) So: \( h = d \) --- **Step 4: Use the fact that both wires reach the same height** From the first wire: \( h = 50\sqrt{3} \) From the second wire: \( h = d \) So \( d = 50\sqrt{3} \). This is consistent — the second wire’s anchor is farther away because of the smaller angle. --- **Step 5: Conclusion** The height \( h \) is determined entirely by the first wire’s data: \( h = 50\sqrt{3} \) meters. --- **Final answer:** Height = \( 50\sqrt{3} \) meters.

3. Solve: 2cos²(x) - 5cos(x) - 3 = 0 for x ∈ [0, 2π] Answer: x = 2π/3, 4π/3 Solution: Let u = cos(x). The equation becomes 2u² - 5u - 3 = 0. Factor the quadratic: (2u + 1)(u - 3) = 0.
   Full step-by-step solution

   Step 1: Let u = cos(x). The equation becomes 2u² - 5u - 3 = 0. Step 2: Factor the quadratic: (2u + 1)(u - 3) = 0. Step 3: Set each factor to zero: 2u + 1 = 0 or u - 3 = 0. Step 4: Solve for u: u = -1/2 or u = 3. Step 5: Since u = cos(x), we have cos(x) = -1/2 or cos(x) = 3. Step 6: cos(x) = 3 has no solution because cosine only outputs values between -1 and 1. Step 7: Solve cos(x) = -1/2 in [0, 2π]. The reference angle is π/3. Cosine is negative in quadrants II and III. Step 8: In quadrant II: x = π - π/3 = 2π/3. Step 9: In quadrant III: x = π + π/3 = 4π/3. Step 10: The solutions are x = 2π/3 and x = 4π/3. The answer is x = 2π/3, 4π/3.

4. Aroha is an acoustic engineer tuning a noise-cancellation system for a large auditorium. The system uses two microphones that capture sound waves, and the combined signal is modeled by the equation 9cos²θ - 12cosθ + 4 = 0, where θ represents the phase shift between the two microphones in radians, measured from 0 to 2π. Find all possible phase shifts θ in the interval [0, 2π) that make the combined signal zero. Answer: θ = arccos(2/3), 2π - arccos(2/3) Solution: Start with the equation 9cos²θ - 12cosθ + 4 = 0. Recognize this as a perfect square trinomial. Check if (3cosθ - 2)² expands to 9cos²θ - 12cosθ + 4: (3cosθ - 2)² = (3cosθ)² - 2(3cosθ)(2) + 2² = 9cos²θ - 12cosθ + 4.
   Full step-by-step solution

   Step 1: Start with the equation 9cos²θ - 12cosθ + 4 = 0. Step 2: Recognize this as a perfect square trinomial. Check if (3cosθ - 2)² expands to 9cos²θ - 12cosθ + 4: (3cosθ - 2)² = (3cosθ)² - 2(3cosθ)(2) + 2² = 9cos²θ - 12cosθ + 4. Yes, it matches. Step 3: Factor the equation: (3cosθ - 2)² = 0. Step 4: Take the square root of both sides: 3cosθ - 2 = 0. Step 5: Solve for cosθ: 3cosθ = 2, so cosθ = 2/3. Step 6: Find the reference angle: θ_ref = arccos(2/3). Since 2/3 is positive, cosine is positive in quadrants I and IV. Step 7: In quadrant I: θ = arccos(2/3). Step 8: In quadrant IV: θ = 2π - arccos(2/3). Step 9: Both solutions are in the interval [0, 2π). The answer is θ = arccos(2/3) and θ = 2π - arccos(2/3).

5. Liam is designing a suspension bridge where the main cable forms a parabolic shape. The height of the cable above the roadway can be modeled by the function h(x) = 50 - 40cos(πx/100), where x is the horizontal distance in meters from the left tower. At what horizontal distances from the left tower will the cable be exactly 30 meters above the roadway? Answer: x = 100/3 meters and x = 200/3 meters Solution: Trigonometric equations often model periodic phenomena like waves, vibrations, or in this case, bridge cables. For cosine equations, we use the property that cos(θ) = cos(-θ) to find symmetric solutions, then apply the period to find all valid answers.
   Full step-by-step solution

   Trigonometric equations often model periodic phenomena like waves, vibrations, or in this case, bridge cables. To solve such equations, we typically isolate the trigonometric function, then use inverse trigonometric functions to find principal solutions. Since trigonometric functions are periodic, we must consider all angles that satisfy the equation within the given domain. For cosine equations, we use the property that cos(θ) = cos(-θ) to find symmetric solutions, then apply the period to find all valid answers.

6. Emma is an astronomer analyzing the light intensity from a distant star. The star's brightness fluctuates due to an orbiting exoplanet, and the relative intensity is modeled by the equation 10 cos² θ − 5 cos θ − 5 = 0, where θ represents the orbital phase angle in radians, measured from 0 to 2π. Find all orbital phase angles θ in the interval [0, 2π) that make the relative intensity zero. Answer: θ = 0, 2π/3, 4π/3 Solution: Write the equation: 10 cos² θ − 5 cos θ − 5 = 0. Factor out the common factor of 5: 5(2 cos² θ − cos θ − 1) = 0. Divide both sides by 5: 2 cos² θ − cos θ − 1 = 0.
   Full step-by-step solution

   Step 1: Write the equation: 10 cos² θ − 5 cos θ − 5 = 0. Step 2: Factor out the common factor of 5: 5(2 cos² θ − cos θ − 1) = 0. Step 3: Divide both sides by 5: 2 cos² θ − cos θ − 1 = 0. Step 4: Factor the quadratic in terms of cos θ. We need two numbers that multiply to 2 × (−1) = −2 and add to −1. These numbers are −2 and 1. Step 5: Rewrite: 2 cos² θ − 2 cos θ + cos θ − 1 = 0. Step 6: Factor by grouping: 2 cos θ (cos θ − 1) + 1 (cos θ − 1) = 0. Step 7: Factor out (cos θ − 1): (cos θ − 1)(2 cos θ + 1) = 0. Step 8: Set each factor to zero: Case 1: cos θ − 1 = 0 → cos θ = 1. Case 2: 2 cos θ + 1 = 0 → cos θ = −1/2. Step 9: Solve Case 1: cos θ = 1. In [0, 2π), the only solution is θ = 0. Step 10: Solve Case 2: cos θ = −1/2. Cosine is negative in quadrants II and III. The reference angle is arccos(1/2) = π/3. So: θ = π − π/3 = 2π/3 (quadrant II) θ = π + π/3 = 4π/3 (quadrant III) Step 11: All solutions in [0, 2π) are θ = 0, 2π/3, 4π/3. The answer is θ = 0, 2π/3, 4π/3.

7. Solve: 2cos²(x) - 5cos(x) + 2 = 0 for x ∈ [0, 2π] Answer: x = π/3, 5π/3 Solution: Let u = cos(x). The equation becomes 2u² - 5u + 2 = 0. Factor the quadratic: (2u - 1)(u - 2) = 0.
   Full step-by-step solution

   Step 1: Let u = cos(x). The equation becomes 2u² - 5u + 2 = 0. Step 2: Factor the quadratic: (2u - 1)(u - 2) = 0. Step 3: Solve for u: 2u - 1 = 0 gives u = 1/2; u - 2 = 0 gives u = 2. Step 4: Since u = cos(x), we have cos(x) = 1/2 or cos(x) = 2. Step 5: cos(x) = 2 has no solution because the cosine function only outputs values between -1 and 1. Step 6: Solve cos(x) = 1/2 in the interval [0, 2π]. Step 7: The reference angle is π/3. Cosine is positive in quadrants I and IV. Step 8: In quadrant I: x = π/3. Step 9: In quadrant IV: x = 2π - π/3 = 5π/3. Step 10: The solutions are x = π/3 and x = 5π/3. The answer is x = π/3, 5π/3.