Trigonometric Equations and Inequalities - AP Precalculus

1. Solve $\sin(x)=\frac{1}{2}$ for $x$ in $[0,2\pi]$ using reference angles.​
   Answer: $x=\left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}$ 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.
2. The equation $\sin(x) = \cos(2x)$ has how many solutions on the interval $[0, 2\pi)$?
   Answer: 3 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.
3. What is the solution set to the inequality $\tan(\theta) < -1$ on the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$?
   Answer: $(-\frac{\pi}{2}, -\frac{\pi}{4})$ The graph of $y = \tan(\theta)$ has vertical asymptotes at $\theta = \pm \frac{\pi}{2}$. The function is strictly increasing on the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. The equation $\tan(\theta) = -1$ has the solution $\theta = -\frac{\pi}{4}$ in this interval. Since the tangent function is increasing, the values of $\theta$ for which $\tan(\theta) < -1$ must be less than $-\frac{\pi}{4}$. The interval is bounded on the left by the asymptote at $\theta = -\frac{\pi}{2}$. Thus, the solution set is $(-\frac{\pi}{2}, -\frac{\pi}{4})$.
4. What are the solutions for $\sqrt{3}\sin(x) = \cos(x)$ on the interval $[0, 2\pi]$?
   Answer: $x = \frac{\pi}{6}, \frac{7\pi}{6}$ 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.
5. What are all solutions to $3\tan^2(x) - 1 = 0$ on the interval $[0, \pi]$?
   Answer: $x = \frac{\pi}{6}, \frac{5\pi}{6}$ The equation is equivalent to $\tan^2(x) = \frac{1}{3}$. Taking the square root of both sides, we get $\tan(x) = \pm \frac{1}{\sqrt{3}}$. For $\tan(x) = \frac{1}{\sqrt{3}}$, the solution in $[0, \pi]$ is $x = \frac{\pi}{6}$. For $\tan(x) = -\frac{1}{\sqrt{3}}$, the solution in $[0, \pi]$ is $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$. Thus, the solutions are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.
6. How many distinct solutions does the equation $2\sin(3x) = \sqrt{2}$ have in the interval $[0, 2\pi)$?
   Answer: 6 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.
7. Find all solutions to the equation $2\cos^2(\theta) + 3\sin(\theta) - 3 = 0$ on the interval $[0, 2\pi]$.
   Answer: $\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}$ Use the Pythagorean identity $\cos^2(\theta) = 1 - \sin^2(\theta)$. The equation becomes $2(1 - \sin^2(\theta)) + 3\sin(\theta) - 3 = 0$, which simplifies to $-2\sin^2(\theta) + 3\sin(\theta) - 1 = 0$, or $2\sin^2(\theta) - 3\sin(\theta) + 1 = 0$. Factoring the quadratic in terms of $\sin(\theta)$ gives $(2\sin(\theta) - 1)(\sin(\theta) - 1) = 0$. So, $\sin(\theta) = \frac{1}{2}$ or $\sin(\theta) = 1$. The solutions for $\sin(\theta) = \frac{1}{2}$ on $[0, 2\pi]$ are $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$. The solution for $\sin(\theta) = 1$ is $\theta = \frac{\pi}{2}$.
8. What are all solutions to the equation $\tan(x) + 1 = 0$ on the interval $[-\pi, \pi]$?
   Answer: $x = \frac{3\pi}{4}, -\frac{\pi}{4}$ The equation is equivalent to $\tan(x) = -1$. The reference angle for which $\tan(x)=1$ is $\frac{\pi}{4}$. The tangent function is negative in Quadrants II and IV. The general solution is $x = \frac{3\pi}{4} + k\pi$ for any integer $k$. For $k=0$, we get $x=\frac{3\pi}{4}$. For $k=-1$, we get $x = \frac{3\pi}{4} - \pi = -\frac{\pi}{4}$. Both of these solutions are in the interval $[-\pi, \pi]$.
9. Which of the following are the solutions to $\tan^2(x) - \sec(x) - 1 = 0$ on the interval $[0, 2\pi)$?
   Answer: $x = \frac{\pi}{3}, \pi, \frac{5\pi}{3}$ Using the Pythagorean identity $\tan^2(x) = \sec^2(x) - 1$, the equation becomes $(\sec^2(x) - 1) - \sec(x) - 1 = 0$, which simplifies to $\sec^2(x) - \sec(x) - 2 = 0$. Factoring gives $(\sec(x) - 2)(\sec(x) + 1) = 0$. Thus, $\sec(x) = 2$ or $\sec(x) = -1$. This is equivalent to $\cos(x) = \frac{1}{2}$ or $\cos(x) = -1$. For $\cos(x) = \frac{1}{2}$, $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$. For $\cos(x) = -1$, $x = \pi$. All three solutions are in the interval $[0, 2\pi)$.
10. Find $\theta$ if $\sin(\theta)=-\frac{\sqrt{2}}{2}$ and $0\le\theta<2\pi$.​
    Answer: $\theta=\left\{\frac{5\pi}{4},\frac{7\pi}{4}\right\}$ 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.
11. What are the solutions to the equation $2\sin(x) + \sqrt{3} = 0$ on the interval $[0, 2\pi]$?
    Answer: $x = \frac{4\pi}{3}, \frac{5\pi}{3}$ 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 $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$.
12. What is the solution set for the inequality $4\sin^2(x) < 3$ on the interval $[0, \pi]$?
    Answer: $[0, \frac{\pi}{3}) \cup (\frac{2\pi}{3}, \pi]$ 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 $(\frac{2\pi}{3}, \pi]$.
13. For what values of $x$ in the interval $[0, 2\pi]$ is the inequality $\sqrt{2}\cos(x) + 1 \le 0$ satisfied?
    Answer: $[\frac{3\pi}{4}, \frac{5\pi}{4}]$ The inequality is equivalent to $\cos(x) \le -\frac{1}{\sqrt{2}}$ or $\cos(x) \le -\frac{\sqrt{2}}{2}$. The values of $x$ for which $\cos(x) = -\frac{\sqrt{2}}{2}$ are $x = \frac{3\pi}{4}$ and $x = \frac{5\pi}{4}$. On the unit circle, the x-coordinate represents the cosine of the angle. We are looking for angles where the x-coordinate is less than or equal to $-\frac{\sqrt{2}}{2}$. This occurs for angles between $\frac{3\pi}{4}$ and $\frac{5\pi}{4}$, inclusive. Therefore, the solution is the interval $[\frac{3\pi}{4}, \frac{5\pi}{4}]$.
14. Find $\theta$ in $[0,\pi]$ if \cos(\theta)=\frac{\sqrt{3}}{2}; use $\theta=\cos^{-1}\!\left(\frac{\sqrt{3}}{2}\right)$.
    Answer: $\theta=\frac{\pi}{6}$ 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.
15. Which of the following is a solution to the equation $\sin(2x)=0$?
    Answer: $x=\frac{\pi}{2}$ 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.
16. Solve $2\cos^2(x)-1=0$ on $[0,2\pi)$ using $\cos(2x)=2\cos^2(x)-1$.
    Answer: $x=\left\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\right\}$ 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.
17. What are the solutions to $2\cos(\theta - \frac{\pi}{6}) = \sqrt{3}$ on the interval $[0, 2\pi]$?
    Answer: $\theta = 0, \frac{\pi}{3}, 2\pi$ 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 $[0, 2\pi]$.
18. Which of the following represents all solutions to the equation $2\cos(\theta) + \sqrt{2} = 0$? Let $k$ be any integer.
    Answer: $\theta = \frac{3\pi}{4} + 2k\pi$ and $\theta = \frac{5\pi}{4} + 2k\pi$ 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 $\theta = \frac{5\pi}{4} + 2k\pi$.
19. On which of the following days is the number of hours of daylight approximately 14.5 hours?
    Answer: Day 171 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.
20. What is the general solution to the equation $4\csc^2(\theta) - 8 = 0$? Let $k$ be any integer.
    Answer: $\theta = \frac{\pi}{4} + \frac{k\pi}{2}$ 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 $\theta = \frac{\pi}{4} + \frac{k\pi}{2}$.

Start this quiz