\begin{definition term="Trigonometric Equation">An equation that involves trigonometric functions of an unknown angle.
\begin{callout type="note">Trigonometric equations often have multiple solutions within a given interval. \end{callout}
Solving Trigonometric Equations
Using the Unit Circle
The unit circle is a powerful tool for solving trigonometric equations because it visually represents the values of sine, cosine, and tangent for angles between \$0^\circ\$ and \$360^\circ\$ (or \$0\$ and \$2\pi\$ radians).
\begin{callout type="example">Solve the equation \$\sin x = 0.5\$ for \$0^\circ \leq x < 360^\circ\$.
- Identify the Reference Angle: The reference angle for \$\sin x = 0.5\$ is \$30^\circ\$ because \$\sin 30^\circ = 0.5\$.
- Determine the Quadrants: Sine is positive in Quadrants I and II.
- Find the Solutions:
- Quadrant I: \$x = 30^\circ\$
- Quadrant II: \$x = 180^\circ - 30^\circ = 150^\circ\$
- Solution Set: \${30^\circ, 150^\circ}\$ \end{callout}
\begin{callout type="note">The reference angle is the acute angle formed by the terminal side of the angle and the x-axis. \end{callout}
Using Inverse Trigonometric Functions
Inverse trigonometric functions can be used to find the principal value of the angle.
\begin{callout type="example">Solve the equation \$\cos x = -0.5\$ for \$0 \leq x < 2\pi\$.
- Find the Reference Angle: \$\cos^{-1}(0.5) = \frac{\pi}{3}\$.
- Determine the Quadrants: Cosine is negative in Quadrants II and III.
- Find the Solutions:
- Quadrant II: \$x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\$
- Quadrant III: \$x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}\$
- Solution Set: \${\frac{2\pi}{3}, \frac{4\pi}{3}}\$ \end{callout}
\begin{callout type="warning">Inverse trigonometric functions only return the principal value, so you must consider the symmetry of the trigonometric function to find all solutions. \end{callout}
General Solutions
Trigonometric equations often have infinitely many solutions because trigonometric functions are periodic.
\begin{callout type="example">Solve the equation \$\tan x = 1\$ for all \$x\$.
- Find the Principal Solution: \$\tan^{-1}(1) = \frac{\pi}{4}\$.
- Use the Periodicity of Tangent: The period of tangent is \$\pi\$, so the general solution is \$x = \frac{\pi}{4} + n\pi\$, where \$n \in \mathbb{Z}\$. \end{callout}
\begin{callout type="note">The period of a trigonometric function is the smallest positive interval over which the function repeats its values.
- Sine and cosine have a period of \$2\pi\$.
- Tangent and cotangent have a period of \$\pi\$. \end{callout}
Special Cases
No Solutions
Some trigonometric equations have no solutions within a given interval.
\begin{callout type="example">Solve the equation \$\sin x = 2\$ for \$0 \leq x < 2\pi\$.
- The sine function only takes values between \$-1\$ and \$1\$, so there are no solutions. \end{callout}
One Solution
Some equations have exactly one solution.
\begin{callout type="example">Solve the equation \$\cos x = 1\$ for \$0 \leq x < 2\pi\$.
- The cosine function equals \$1\$ only at \$x = 0\$. \end{callout}
\begin{callout type="self_review">1. Solve the equation \$\sin x = -0.5\$ for \$0 \leq x < 2\pi\$. 2. Find the general solution for \$\cos x = \frac{1}{2}\$. 3. Determine if the equation \$\tan x = 2\$ has solutions in the interval [[0, \pi]]. \end{callout}
\begin{callout type="tok">How does the periodic nature of trigonometric functions challenge our understanding of "solving" an equation? In what ways does this differ from solving algebraic equations? \end{callout}
