AP PreCalculus Practice Quiz: Trigonometric Equations and Inequalities

Correct Answer: A

First, solve for sin(x): 2sin(x) = 1, so sin(x) = 1/2. The sine function is positive in Quadrants I and II. The reference angle for sin(x) = 1/2 is π/6. The solution in Quadrant I is x = π/6. The solution in Quadrant II is x = π - π/6 = 5π/6. Both solutions are within the specified interval [0, 2π).

Correct Answer: B

The tangent function has a period of π. The principal value solution for arctan(-1) is -π/4. A coterminal angle in the interval [0, 2π) is 3π/4 (in Quadrant II). Since the period of tangent is π, all solutions are separated by integer multiples of π. Therefore, the general solution is x = 3π/4 + nπ, which covers all solutions in both Quadrant II and Quadrant IV.

Correct Answer: C

Set D(t) = 6.5 to get 3cos(πt/6) + 5 = 6.5. This simplifies to 3cos(πt/6) = 1.5, or cos(πt/6) = 0.5. Let θ = πt/6. The solutions for cos(θ) = 0.5 are θ = π/3 + 2nπ and θ = 5π/3 + 2nπ. Substitute back for t: (1) πt/6 = π/3 + 2nπ gives t = 2 + 12n. In the interval [0, 24], this yields t=2 (for n=0) and t=14 (for n=1). (2) πt/6 = 5π/3 + 2nπ gives t = 10 + 12n. In the interval [0, 24], this yields t=10 (for n=0) and t=22 (for n=1). In total, there are 4 solutions.

Correct Answer: C

The value x ≈ 2.498 is in Quadrant II, where cosine is negative. Cosine is also negative in Quadrant III. Due to the symmetry of the cosine function graph about the y-axis, if x₀ is a solution, then -x₀ is also a solution. To find a solution within [0, 2π), we can use 2π - x₀. Therefore, the other solution is approximately 2π - 2.498 ≈ 6.283 - 2.498 = 3.785.

Correct Answer: A

First, find the values where sin(x) = √3/2. In the interval [0, 2π), these are x = π/3 and x = 2π/3. By visualizing the graph of y = sin(x), the function's value is above the line y = √3/2 between these two x-values. Therefore, the solution to the inequality is the open interval (π/3, 2π/3).

Correct Answer: D

The equation simplifies to cos(3x) = 1/2. Let θ = 3x. If the domain for x is [0, 2π), then the domain for θ is [0, 6π). We need to find how many times cos(θ) = 1/2 in the interval [0, 6π). The cosine function completes 3 full periods in this interval. In each period [0, 2π), there are two solutions (at π/3 and 5π/3). Therefore, over three periods, there will be 3 * 2 = 6 solutions for θ, which correspond to 6 unique solutions for x in [0, 2π).

Correct Answer: B

The fundamental reason for infinitely many solutions is the periodic nature of trigonometric functions. The graph of y = sin(x) repeats its values every 2π units. Therefore, if x₀ is a solution, then x₀ + 2nπ (for any integer n) will also be a solution, leading to an infinite set of solutions.

Correct Answer: A

This equation is quadratic in form. Let u = cos(x). The equation becomes 2u² + u - 1 = 0. Factoring gives (2u - 1)(u + 1) = 0. This yields two possibilities: u = 1/2 or u = -1. Case 1: cos(x) = 1/2. In [0, 2π), the solutions are x = π/3 and x = 5π/3. Case 2: cos(x) = -1. In [0, 2π), the solution is x = π. Combining all unique solutions gives {π/3, π, 5π/3}.

Correct Answer: C

The projectile lands when h(t) = 0 for t > 0. Set h(t) = 0: -5t² + 20t sin(θ) = 0. Factor out 5t: 5t(-t + 4sin(θ)) = 0. This gives t=0 (the launch) or t = 4sin(θ) (the landing). The problem states the landing time is t = 2√3. So, we set 2√3 = 4sin(θ), which simplifies to sin(θ) = (2√3)/4 = √3/2. On the interval (0, π), the solutions to sin(θ) = √3/2 are θ = π/3 and θ = 2π/3. Of the options provided, π/3 is a possible value.

Correct Answer: C

This question highlights that inverse trigonometric functions have restricted domains and may not give all solutions in a desired interval. For sin(x) = -1/2, the sine function is negative in Quadrants III and IV. The reference angle is arcsin(1/2) = π/6. The solution in Quadrant III is π + π/6 = 7π/6. The solution in Quadrant IV is 2π - π/6 = 11π/6. The principal value -π/6 is coterminal with 11π/6, but 7π/6 is the other distinct solution in the interval.