Trigonometric Equations and Inequalities - AP Precalculus

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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