AP Precalculus Flashcards: Trigonometric Equations And Inequalities

What this deck covers

This deck focuses on Trigonometric Equations And Inequalities, giving you a quick way to review the definitions, rules, and examples that matter most for AP Precalculus.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Solve $\text{tan}(\theta) = \frac{1}{\sqrt{3}}$ in $[0, 2\pi)$.

Answer: $\theta = \frac{\pi}{6}, \frac{7\pi}{6}$. Since $\tan(\theta) = \frac{1}{\sqrt{3}}$, we have $\theta = \frac{\pi}{6}$ and $\frac{7\pi}{6}$.

Flashcard 2: Find the solutions for $\cot(\theta) = 0$ in $[0, 2\pi)$.

Answer: $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$. Cotangent equals zero when cosine is zero.

Flashcard 3: State the double angle formula for $\text{cos}(2\theta)$.

Answer: $\text{cos}(2\theta) = \text{cos}^2(\theta) - \text{sin}^2(\theta)$. This is the standard double angle formula for cosine.

Flashcard 4: Solve $\text{sin}(2\theta) = 0$ for $\theta$ in $[0, 2\text{π})$.

Answer: $\theta = 0, \frac{\text{π}}{2}, \text{π}, \frac{3\text{π}}{2}$. When $\sin(2\theta) = 0$, then $2\theta = n\text{π}$, so $\theta = \frac{n\text{π}}{2}$.

Flashcard 5: What is the period of $\text{cos}(2\theta)$?

Answer: $\text{π}$. Period of $\cos(b\theta)$ is $\frac{2\text{π}}{b}$, so $\frac{2\text{π}}{2} = \text{π}$.

Flashcard 6: What is the general solution for $\text{csc}(\theta) = a$?

Answer: $\theta = \text{csc}^{-1}(a) + 2n\text{π} \text{ or } -\text{csc}^{-1}(a) + 2n\text{π}$. Cosecant is symmetric, giving two solutions per period.

Flashcard 7: Find the solutions for $\text{cot}(\theta) = 0$ in $[0, 2\text{π})$.

Answer: $\theta = \frac{\text{π}}{2}, \frac{3\text{π}}{2}$. Cotangent equals zero when cosine is zero.

Flashcard 8: Solve $\text{cos}^2(\theta) = \frac{1}{2}$ for $\theta$ in $[0, 2\text{π})$.

Answer: $\theta = \frac{\text{π}}{4}, \frac{3\text{π}}{4}, \frac{5\text{π}}{4}, \frac{7\text{π}}{4}$. Taking square root gives $\cos(\theta) = \pm\frac{\sqrt{2}}{2}$, yielding four solutions.

Flashcard 9: What is the general solution for $\text{sec}(\theta) = a$?

Answer: $\theta = \text{sec}^{-1}(a) + 2n\text{π} \text{ or } -\text{sec}^{-1}(a) + 2n\text{π}$. Secant is symmetric about the y-axis, giving two solutions per period.

Flashcard 10: Find the solutions for $\text{sin}(\theta) = 1$ in $[0, 2\text{π})$.

Answer: $\theta = \frac{\text{π}}{2}$. Sine equals 1 only at $\frac{\text{π}}{2}$ in the interval $[0, 2\text{π})$.

Flashcard 11: What is the period of $\tan(3\theta)$?

Answer: $\frac{\text{π}}{3}$. Period of $\tan(b\theta)$ is $\frac{\text{π}}{b}$, so $\frac{\text{π}}{3}$ for $b=3$.

Flashcard 12: What is the period of $\text{csc}(\theta)$?

Answer: $2\text{π}$. Cosecant has the same period as sine, which is $2\text{π}$.

Flashcard 13: Find $\theta$ if $\text{sec}(\theta) = 2$ in $[0, 2\text{π})$.

Answer: $\theta = \frac{\text{π}}{3}, \frac{5\text{π}}{3}$. Since $\sec(\theta) = \frac{1}{\cos(\theta)}$, we need $\cos(\theta) = \frac{1}{2}$.

Flashcard 14: Solve $\text{cos}(\theta) = \frac{1}{2}$ in $[0, 2\text{π})$.

Answer: $\theta = \frac{\text{π}}{3}, \frac{5\text{π}}{3}$. Cosine equals $\frac{1}{2}$ at angles $\frac{\text{π}}{3}$ and $\frac{5\text{π}}{3}$ in $[0, 2\text{π})$.

Flashcard 15: Solve $\tan(\theta) = \frac{1}{\sqrt{3}}$ in $[0, 2\pi)$.

Answer: $\theta = \frac{\pi}{6}, \frac{7\pi}{6}$. Since $\tan(\theta) = \frac{1}{\sqrt{3}}$, we have $\theta = \frac{\pi}{6}$ and $\frac{7\pi}{6}$.

Flashcard 16: Find the solutions for $\text{sin}(\theta) = -\frac{\text{√2}}{2}$ in $[0, 2\text{π})$.

Answer: $\theta = \frac{5\text{π}}{4}, \frac{7\text{π}}{4}$. Sine equals $-\frac{\sqrt{2}}{2}$ in the third and fourth quadrants.

Flashcard 17: What is the period of $\text{cot}(\theta)$?

Answer: $\text{π}$. Cotangent function repeats its values every $\text{π}$ radians.

Flashcard 18: What is the period of $\text{csc}(2\theta)$?

Answer: $\text{π}$. Period of $\csc(b\theta)$ is $\frac{2\text{π}}{b}$, so $\frac{2\text{π}}{2} = \text{π}$.

Flashcard 19: What is the general solution for $\text{sin}(\theta) = a$?

Answer: $\theta = \text{sin}^{-1}(a) + 2n\text{π} \text{ or } \text{π} - \text{sin}^{-1}(a) + 2n\text{π}$. Sine has two solutions per period due to its symmetry properties.

Flashcard 20: What is the period of $\text{sin}(4\theta)$?

Answer: $\frac{\text{π}}{2}$. Period of $\sin(b\theta)$ is $\frac{2\text{π}}{b}$, so $\frac{2\text{π}}{4} = \frac{\text{π}}{2}$.

Flashcard 21: What is the period of $\tan(\theta)$?

Answer: $\text{π}$. Tangent function repeats its values every $\text{π}$ radians.

Flashcard 22: State the identity for $\text{csc}(\theta)$ in terms of $\text{sin}(\theta)$.

Answer: $\text{csc}(\theta) = \frac{1}{\text{sin}(\theta)}$. Cosecant is the reciprocal of sine.

Flashcard 23: Solve $\sin(\theta) = \frac{\sqrt{3}}{2}$ in $[0, 2\pi)$.

Answer: $\theta = \frac{\pi}{3}, \frac{2\pi}{3}$. Sine equals $\frac{\sqrt{3}}{2}$ at angles $\frac{\pi}{3}$ and $\frac{2\pi}{3}$.

Flashcard 24: State the Pythagorean identity for $\text{tan}^2(\theta)$.

Answer: $1 + \tan^2(\theta) = \text{sec}^2(\theta)$. This is the fundamental Pythagorean identity involving tangent and secant.

Flashcard 25: What is the period of $\text{tan}(4\theta)$?

Answer: $\frac{\text{π}}{4}$. Period of $\tan(b\theta)$ is $\frac{\text{π}}{b}$, so $\frac{\text{π}}{4}$ for $b=4$.

Flashcard 26: Find the solutions for $\text{cos}(\theta) = -\frac{1}{2}$ in $[0, 2\text{π})$.

Answer: $\theta = \frac{2\text{π}}{3}, \frac{4\text{π}}{3}$. Cosine equals $-\frac{1}{2}$ at angles $\frac{2\text{π}}{3}$ and $\frac{4\text{π}}{3}$.

Flashcard 27: What is the period of $\text{cot}(3\theta)$?

Answer: $\frac{\text{π}}{3}$. Period of $\cot(b\theta)$ is $\frac{\text{π}}{b}$, so $\frac{\text{π}}{3}$ for $b=3$.

Flashcard 28: What is the general solution for $\text{cos}(\theta) = a$?

Answer: $\theta = \text{cos}^{-1}(a) + 2n\text{π} \text{ or } -\text{cos}^{-1}(a) + 2n\text{π}$. Cosine is symmetric about the y-axis, giving two solutions per period.

Flashcard 29: Find the solutions for $\text{tan}(\theta) = 1$ in $[0, 2\text{π})$.

Answer: $\theta = \frac{\text{π}}{4}, \frac{5\text{π}}{4}$. Tangent equals 1 at $\frac{\text{π}}{4}$ and repeats every $\text{π}$.

Flashcard 30: Find the solutions for $\text{csc}(\theta) = -2$ in $[0, 2\text{π})$.

Answer: $\theta = \frac{7\text{π}}{6}, \frac{11\text{π}}{6}$. Since $\csc(\theta) = -2$, we need $\sin(\theta) = -\frac{1}{2}$.

Flashcard 31: What is the period of $\text{cos}(4\theta)$?

Answer: $\frac{\text{π}}{2}$. Period of $\cos(b\theta)$ is $\frac{2\text{π}}{b}$, so $\frac{2\text{π}}{4} = \frac{\text{π}}{2}$.

Flashcard 32: What is the amplitude of $\text{sin}(3\theta)$?

Answer:

1. Amplitude is the coefficient of the sine function, which is 1.

Flashcard 33: What is the general solution for $\text{sin}(\theta) = 0$?

Answer: $\theta = n\text{π}$, $n \text{ is an integer}$. Sine equals zero at multiples of $\text{π}$.

Flashcard 34: Find $\theta$ if $\sec(\theta) = 2$ in $[0, 2\pi)$.

Answer: $\theta = \frac{\pi}{3}, \frac{5\pi}{3}$. Since $\sec(\theta) = \frac{1}{\cos(\theta)}$, we need $\cos(\theta) = \frac{1}{2}$.

Flashcard 35: What is the period of $\text{sec}(\theta)$?

Answer: $2\text{π}$. Secant has the same period as cosine, which is $2\text{π}$.

Flashcard 36: What is the period of $\text{sin}(\theta)$?

Answer: $2\text{π}$. Sine function completes one cycle every $2\text{π}$ radians.

Flashcard 37: State the identity for $\text{sec}(\theta)$ in terms of $\text{cos}(\theta)$.

Answer: $\text{sec}(\theta) = \frac{1}{\text{cos}(\theta)}$. Secant is the reciprocal of cosine.

Flashcard 38: Find the solutions for $\text{cos}(\theta) = 0$ in $[0, 2\text{π})$.

Answer: $\theta = \frac{\text{π}}{2}, \frac{3\text{π}}{2}$. Cosine equals zero when the angle is an odd multiple of $\frac{\text{π}}{2}$.

Flashcard 39: Find the solutions for $\text{sec}(\theta) = -2$ in $[0, 2\text{π})$.

Answer: $\theta = \frac{2\text{π}}{3}, \frac{4\text{π}}{3}$. Since $\sec(\theta) = -2$, we need $\cos(\theta) = -\frac{1}{2}$.