AP Precalculus Quiz: Trigonometric Equations And Inequalities

What this quiz covers

This quiz focuses on Trigonometric Equations And Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Day 80
2. Day 125
3. Day 171 (correct answer)
4. Day 263

Explanation: Set $D(t) = 14.5$ to solve for $t$: $14.5 = 12 + 2.5\sin(\frac{2\pi}{365}(t-80))$. Subtracting 12 gives $2.5 = 2.5\sin(\frac{2\pi}{365}(t-80))$, which simplifies to $1 = \sin(\frac{2\pi}{365}(t-80))$. The principal value for which sine is 1 is $\frac{\pi}{2}$. So, $\frac{2\pi}{365}(t-80) = \frac{\pi}{2}$. Dividing by $2\pi$ gives $\frac{1}{365}(t-80) = \frac{1}{4}$. Thus, $t-80 = \frac{365}{4} = 91.25$. Solving for $t$ gives $t = 171.25$. This corresponds to day 171.

Question 2

1. $x=\left\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\right\}$ (correct answer)
2. $x=\left\{\frac{\pi}{2},\pi,\frac{3\pi}{2}\right\}$
3. $x=\left\{\frac{\pi}{8},\frac{5\pi}{8},\frac{9\pi}{8},\frac{13\pi}{8}\right\}$
4. $x=\left\{\frac{\pi}{4},\frac{3\pi}{4}\right\}$

Explanation: This question tests AP Precalculus skills, specifically solving equations involving composite arguments. The equation cos(2x) = 0 means 2x must equal π/2 + kπ for integer k, since cosine equals zero at odd multiples of π/2. Solving for x gives x = π/4 + kπ/2, and within [0, 2π) we get x = π/4, 3π/4, 5π/4, and 7π/4. Choice A is correct because it lists all four solutions obtained by setting k = 0, 1, 2, 3 in the formula x = π/4 + kπ/2. Choice C is incorrect because it incorrectly divides the angles by 2 again, suggesting x = π/8 + kπ/4, a common error when handling composite arguments. To help students: Emphasize solving for the composite argument first (2x) before solving for x. Practice systematic enumeration of solutions within given intervals.

Question 3

1. $x=\left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}$ (correct answer)
2. $x=\left\{\frac{\pi}{3},\frac{2\pi}{3}\right\}$
3. $x=\left\{\frac{\pi}{6}\right\}$
4. $x=\left\{30^\circ,150^\circ\right\}$

Explanation: This question tests AP Precalculus skills, specifically solving basic trigonometric equations using reference angles. The equation sin(x) = 1/2 requires finding all angles in [0, 2π] where sine equals one-half. Using the reference angle π/6 (where sin(π/6) = 1/2), we find solutions in quadrants where sine is positive (I and II). Choice A is correct because sin(π/6) = sin(5π/6) = 1/2, giving both solutions in the specified interval. Choice D gives the same angles in degrees (30° and 150°), but the interval notation [0, 2π] indicates radians are required. To help students: Emphasize using reference angles to find all solutions systematically. Practice identifying which quadrants have positive or negative values for each trigonometric function.

Question 4

1. $x=\frac{\pi}{4}$
2. $x=\frac{\pi}{3}$
3. $x=\frac{\pi}{2}$ (correct answer)
4. $x=\frac{3\pi}{4}$

Explanation: The equation $\sin(\theta)=0$ is satisfied when $\theta$ is an integer multiple of $\pi$ (i.e., $\theta = k\pi$). In this problem, $\theta = 2x$. So we need to solve $2x=k\pi$ for an integer $k$. This means $x = \frac{k\pi}{2}$. We check the answer choices. For choice C, if $x=\frac{\pi}{2}$, this corresponds to $k=1$, and it is a valid solution.

Question 5

1. $x = \frac{2\pi}{3}, \frac{4\pi}{3}$
2. $x = \frac{\pi}{3}, \frac{5\pi}{3}$
3. $x = \frac{4\pi}{3}, \frac{5\pi}{3}$ (correct answer)
4. $x = \frac{\pi}{3}, \frac{2\pi}{3}$

Explanation: The equation can be rewritten as $\sin(x) = -\frac{\sqrt{3}}{2}$. The reference angle for which $\sin(x) = \frac{\sqrt{3}}{2}$ is $\frac{\pi}{3}$. Since the sine function is negative in Quadrants III and IV, the solutions on the interval $[0, 2\pi]$ are $x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$ and $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$.

Question 6

1. $x = \frac{\pi}{3}, \frac{4\pi}{3}$
2. $x = \frac{2\pi}{3}, \frac{5\pi}{3}$
3. $x = \frac{\pi}{6}, \frac{7\pi}{6}$ (correct answer)
4. $x = \frac{5\pi}{6}, \frac{11\pi}{6}$

Explanation: If $\cos(x) \neq 0$, we can divide both sides by $\cos(x)$ to get $\sqrt{3}\tan(x) = 1$, which means $\tan(x) = \frac{1}{\sqrt{3}}$. The reference angle is $\frac{\pi}{6}$. Since tangent is positive in Quadrants I and III, the solutions are $x = \frac{\pi}{6}$ and $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$. We must check if any solutions were lost by assuming $\cos(x) \neq 0$. If $\cos(x)=0$, then $x=\frac{\pi}{2}$ or $x=\frac{3\pi}{2}$. In these cases, $\sin(x)$ is $1$ or $-1$. The equation becomes $\sqrt{3}(\pm 1) = 0$, which is false. So no solutions were lost.

Question 7

1. 2
2. 3
3. 4
4. 6 (correct answer)

Explanation: The equation is $\sin(3x) = \frac{\sqrt{2}}{2}$. Let $u = 3x$. Since $0 \le x < 2\pi$, we have $0 \le 3x < 6\pi$, so $0 \le u < 6\pi$. The solutions for $\sin(u) = \frac{\sqrt{2}}{2}$ in $[0, 2\pi)$ are $u = \frac{\pi}{4}$ and $u = \frac{3\pi}{4}$. To find all solutions in $[0, 6\pi)$, we add multiples of $2\pi$. The solutions for $u$ are: $\frac{\pi}{4}$, $\frac{3\pi}{4}$, $\frac{\pi}{4}+2\pi=\frac{9\pi}{4}$, $\frac{3\pi}{4}+2\pi=\frac{11\pi}{4}$, $\frac{\pi}{4}+4\pi=\frac{17\pi}{4}$, and $\frac{3\pi}{4}+4\pi=\frac{19\pi}{4}$. Each of these six values of $u$ gives a distinct value of $x = u/3$ in the interval $[0, 2\pi)$. Therefore, there are 6 solutions.

Question 8

1. $\theta = \frac{\pi}{4} + 2k\pi$ and $\theta = \frac{3\pi}{4} + 2k\pi$
2. $\theta = \frac{\pi}{4} + k\pi$
3. $\theta = \frac{\pi}{2} + k\pi$
4. $\theta = \frac{\pi}{4} + \frac{k\pi}{2}$ (correct answer)

Explanation: The equation simplifies to $\csc^2(\theta) = 2$. Taking the reciprocal of both sides gives $\sin^2(\theta) = \frac{1}{2}$. Taking the square root of both sides gives $\sin(\theta) = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$. This is true for all angles with a reference angle of $\frac{\pi}{4}$. The solutions in $[0, 2\pi)$ are $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$. These solutions are spaced $\frac{\pi}{2}$ apart. Thus, the general solution can be written compactly as $\theta = \frac{\pi}{4} + \frac{k\pi}{2}$.

Question 9

1. $\theta = \frac{3\pi}{4} + 2k\pi$
2. $\theta = \frac{\pi}{2} + k\pi$
3. $\theta = \frac{3\pi}{4} + k\pi$ (correct answer)
4. $\theta = \frac{\pi}{4} + k\pi$

Explanation: First, isolate the cosine term: $2\cos(2\theta - \frac{\pi}{2}) = -2$, which simplifies to $\cos(2\theta - \frac{\pi}{2}) = -1$. The general solution for $\cos(u) = -1$ is $u = \pi + 2k\pi$. Let $u = 2\theta - \frac{\pi}{2}$. So, $2\theta - \frac{\pi}{2} = \pi + 2k\pi$. Add $\frac{\pi}{2}$ to both sides: $2\theta = \frac{3\pi}{2} + 2k\pi$. Finally, divide by 2: $\theta = \frac{3\pi}{4} + k\pi$.

Question 10

1. $(\frac{\pi}{3}, \frac{2\pi}{3})$
2. $[0, \frac{\pi}{3}) \cup (\frac{2\pi}{3}, \pi]$ (correct answer)
3. $[0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, \pi]$
4. $(\frac{\pi}{6}, \frac{5\pi}{6})$

Explanation: The inequality simplifies to $\sin^2(x) < \frac{3}{4}$, which is equivalent to $-\frac{\sqrt{3}}{2} < \sin(x) < \frac{\sqrt{3}}{2}$. On the interval $[0, \pi]$, $\sin(x)$ is always non-negative, so the inequality becomes $0 \le \sin(x) < \frac{\sqrt{3}}{2}$. The solutions to $\sin(x) = \frac{\sqrt{3}}{2}$ on this interval are $x = \frac{\pi}{3}$ and $x = \frac{2\pi}{3}$. The inequality $0 \le \sin(x) < \frac{\sqrt{3}}{2}$ is satisfied when $x$ is in the interval $[0, \frac{\pi}{3})$ or in the interval $(\frac{2\pi}{3}, \pi]$.

Question 11

1. $\theta=\frac{\pi}{3}$ (correct answer)
2. $\theta=\frac{\pi}{6}$
3. $\theta=\frac{2\pi}{3}$
4. $\theta=60^\circ$

Explanation: This question tests AP Precalculus skills, specifically finding angles with given cosine values. The equation cos(θ) = 1/2 requires identifying angles in the restricted interval [0, π] where cosine equals one-half. On the unit circle, cos(θ) = 1/2 occurs at θ = π/3 (60°) and θ = 5π/3 (300°), but only π/3 lies within [0, π]. Choice A is correct because θ = π/3 is the only solution in the given interval. Choice B gives π/6 where cos(π/6) = √3/2, choice C gives 2π/3 where cos(2π/3) = -1/2, and choice D gives the degree measure instead of radians. To help students: memorize special angle values on the unit circle. Practice converting between degrees and radians, and always check that solutions fall within the specified interval.

Question 12

1. $x=\left\{\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}\right\}$
2. $x=\left\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\right\}$ (correct answer)
3. $x=\left\{\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\right\}$
4. $x=\left\{\frac{\pi}{4},\frac{3\pi}{4}\right\}$

Explanation: This question tests AP Precalculus skills, specifically solving equations using trigonometric identities. The equation 2cos²(x) - 1 = 0 can be recognized as cos(2x) = 0 using the double angle identity cos(2x) = 2cos²(x) - 1. From cos(2x) = 0, we know 2x = π/2 + kπ, giving x = π/4 + kπ/2 for integer k. Choice B is correct because within [0, 2π), this yields x = π/4, 3π/4, 5π/4, and 7π/4. Choice A is incorrect because those angles satisfy sin(2x) = 1/2, not cos(2x) = 0, showing confusion between different trigonometric equations. To help students: Practice recognizing when to use double angle identities in reverse. Verify solutions by substituting back into the original equation.

Question 13

1. $\theta=\left\{\frac{5\pi}{4},\frac{7\pi}{4}\right\}$ (correct answer)
2. $\theta=\left\{\frac{\pi}{4},\frac{3\pi}{4}\right\}$
3. $\theta=\left\{\frac{5\pi}{4}\right\}$
4. $\theta=\left\{-\frac{\pi}{4},-\frac{3\pi}{4}\right\}$

Explanation: This question tests AP Precalculus skills, specifically finding angles with given sine values. The equation sin(θ) = -√2/2 requires identifying angles where sine equals negative square root of 2 over 2. Since sine is negative in quadrants III and IV, and the reference angle is π/4 (where sin(π/4) = √2/2), we find solutions at π + π/4 = 5π/4 and 2π - π/4 = 7π/4. Choice A is correct because sin(5π/4) = sin(7π/4) = -√2/2, giving both angles in [0, 2π) where sine has this value. Choice B incorrectly gives angles where sin(θ) = √2/2 (positive), not the required negative value. To help students: Use the unit circle to identify quadrants where trigonometric functions are negative. Practice finding reference angles and applying them correctly in each quadrant.

Question 14

1. $\theta=\frac{\pi}{6}$ (correct answer)
2. $\theta=\frac{5\pi}{6}$
3. $\theta=\frac{\pi}{3}$
4. $\theta=30^\circ$

Explanation: This question tests AP Precalculus skills, specifically finding angles using inverse cosine within a restricted domain. Given cos(θ) = √3/2, we need to find θ in [0, π] using the inverse cosine function. Since cos(π/6) = √3/2 and π/6 is in the interval [0, π], we have θ = π/6. Choice A is correct because θ = cos⁻¹(√3/2) = π/6, which is the unique solution in the given interval. Choice B (5π/6) is incorrect because cos(5π/6) = -√3/2, not √3/2, showing a common sign error. To help students: Memorize special angle values and their trigonometric ratios. Emphasize that inverse cosine has range [0, π], which matches the given interval, making the solution unique.

Question 15

1. $\theta = 0, \frac{\pi}{3}$
2. $\theta = \frac{\pi}{3}$ only
3. $\theta = 0, \frac{\pi}{3}, 2\pi$ (correct answer)
4. $\theta = \frac{\pi}{6}, \frac{11\pi}{6}$

Explanation: The equation simplifies to $\cos(\theta - \frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. Let $u = \theta - \frac{\pi}{6}$. As $\theta$ ranges from $0$ to $2\pi$, $u$ ranges from $-\frac{\pi}{6}$ to $\frac{11\pi}{6}$. The solutions for $\cos(u) = \frac{\sqrt{3}}{2}$ in this interval for $u$ are $u = -\frac{\pi}{6}$, $u = \frac{\pi}{6}$, and $u = \frac{11\pi}{6}$. Substituting back: 1) $\theta - \frac{\pi}{6} = -\frac{\pi}{6} \implies \theta = 0$. 2) $\theta - \frac{\pi}{6} = \frac{\pi}{6} \implies \theta = \frac{\pi}{3}$. 3) $\theta - \frac{\pi}{6} = \frac{11\pi}{6} \implies \theta = 2\pi$. All three solutions are in the specified interval $[0, 2\pi]$.

Question 16

1. 1
2. 2
3. 3 (correct answer)
4. 4

Explanation: Using the double-angle identity for cosine, $\cos(2x) = 1 - 2\sin^2(x)$, the equation becomes $\sin(x) = 1 - 2\sin^2(x)$. Rearranging gives the quadratic equation $2\sin^2(x) + \sin(x) - 1 = 0$. Factoring this equation yields $(2\sin(x) - 1)(\sin(x) + 1) = 0$. This implies $\sin(x) = \frac{1}{2}$ or $\sin(x) = -1$. On the interval $[0, 2\pi)$, $\sin(x) = \frac{1}{2}$ has two solutions, $x=\frac{\pi}{6}$ and $x=\frac{5\pi}{6}$. On the same interval, $\sin(x) = -1$ has one solution, $x=\frac{3\pi}{2}$. In total, there are 3 distinct solutions.

Question 17

1. $(0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, 2\pi)$
2. $(\frac{7\pi}{6}, \frac{11\pi}{6})$
3. $(\frac{\pi}{6}, \frac{5\pi}{6})$ (correct answer)
4. $(\frac{5\pi}{6}, \frac{7\pi}{6})$

Explanation: The inequality $f(x) > 0$ is equivalent to $2\sin(x)-1 > 0$, or $\sin(x) > \frac{1}{2}$. The solutions to $\sin(x) = \frac{1}{2}$ in $[0, 2\pi]$ are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$. By examining the graph of $y = \sin(x)$ or the unit circle, we see that $\sin(x)$ is greater than $\frac{1}{2}$ for angles strictly between these two values. Therefore, the solution is the interval $(\frac{\pi}{6}, \frac{5\pi}{6})$.

Question 18

1. $\theta=\left\{\frac{\pi}{4},\frac{3\pi}{4}\right\}$
2. $\theta=\left\{\frac{5\pi}{4},\frac{7\pi}{4}\right\}$ (correct answer)
3. $\theta=\left\{\frac{3\pi}{4},\frac{5\pi}{4}\right\}$
4. $\theta=\left\{-\frac{\pi}{4},-\frac{3\pi}{4}\right\}$

Explanation: This question tests AP Precalculus skills, specifically finding angles with negative sine values using reference angles. Given sin(θ) = -√2/2, we need angles where sine has this negative value. The reference angle is π/4 (since sin(π/4) = √2/2), and sine is negative in quadrants III and IV, giving θ = π + π/4 = 5π/4 and θ = 2π - π/4 = 7π/4. Choice B is correct because it identifies both solutions θ = 5π/4 and 7π/4 in [0, 2π). Choice A is incorrect because π/4 and 3π/4 have positive sine values (√2/2), not negative, showing a sign error. To help students: Use the unit circle to visualize where sine is negative (below the x-axis). Apply the reference angle systematically in quadrants III and IV for negative sine values.

Question 19

1. $x\in\left(0,\frac{7\pi}{6}\right)\cup\left(\frac{11\pi}{6},2\pi\right)$ (correct answer)
2. $x\in\left(\frac{7\pi}{6},\frac{11\pi}{6}\right)$
3. $x\in\left[0,\frac{7\pi}{6}\right]\cup\left[\frac{11\pi}{6},2\pi\right)$
4. $x\in\left(0,\frac{5\pi}{6}\right)\cup\left(\frac{7\pi}{6},2\pi\right)$

Explanation: This question tests AP Precalculus skills, specifically solving trigonometric inequalities using the unit circle. The inequality 2sin(x) + 1 > 0 simplifies to sin(x) > -1/2, requiring us to find where sine values exceed -1/2 on the unit circle. On the interval [0, 2π), sin(x) = -1/2 at x = 7π/6 and x = 11π/6, and sine is greater than -1/2 everywhere except between these two values. Choice A is correct because it identifies the solution as (0, 7π/6) ∪ (11π/6, 2π), using open intervals since the inequality is strict. Choice B is incorrect because it gives the complementary interval where sin(x) < -1/2, a common error when misinterpreting inequality directions. To help students: Use the unit circle to visualize where sine values are positive, negative, and equal to key values. Emphasize the difference between strict inequalities (open intervals) and non-strict inequalities (closed intervals).

Question 20

1. $\theta = \frac{3\pi}{4} + 2k\pi$ and $\theta = \frac{5\pi}{4} + 2k\pi$ (correct answer)
2. $\theta = \frac{\pi}{4} + 2k\pi$ and $\theta = \frac{7\pi}{4} + 2k\pi$
3. $\theta = \frac{3\pi}{4} + k\pi$ and $\theta = \frac{5\pi}{4} + k\pi$
4. $\theta = \frac{\pi}{4} + k\pi$ and $\theta = \frac{3\pi}{4} + k\pi$

Explanation: First, solve for $\cos(\theta)$: $\cos(\theta) = -\frac{\sqrt{2}}{2}$. The reference angle is $\frac{\pi}{4}$. The cosine function is negative in Quadrants II and III. The solutions in the interval $[0, 2\pi)$ are $\theta = \frac{3\pi}{4}$ and $\theta = \frac{5\pi}{4}$. Since the cosine function has a period of $2\pi$, the general solution is found by adding integer multiples of $2\pi$ to these base solutions, yielding $\theta = \frac{3\pi}{4} + 2k\pi$ and $\theta = \frac{5\pi}{4} + 2k\pi$.