01 / 20

Solving Trig Equations
in Context

Learn to use inverse trigonometric functions to solve equations from real-world models, find all solutions, and interpret what they mean.

Precalculus

Why Solve Trig Equations in Context?

Trigonometric functions model periodic phenomena: tides, temperatures, sound waves, Ferris wheels, daylight hours, and more. When we ask "at what time does the temperature reach \(75^\circ\text{F}\)?", we're solving a trig equation that arose from a real model.

Key Vocabulary

Inverse Trig Function
\(\sin^{-1}(k)\), \(\cos^{-1}(k)\), \(\tan^{-1}(k)\) give the principal value: one specific angle whose trig ratio equals \(k\).
General Solution
Because trig functions are periodic, there are infinitely many angles that satisfy the equation. We write families like \(\theta + 2\pi n\).
Modeling Context
The real-world scenario that restricts which solutions make sense: a valid time, month, or angle within a physical range.

Visualizing Two Solutions per Period

For \(\sin(\theta) = 0.5\), the unit circle shows two angles in \([0, 2\pi)\): one in Quadrant I and one in Quadrant II. Inverse sine returns only the Q I value.

\(y=0.5\) \(\frac{\pi}{6}\) \(\frac{5\pi}{6}\) Unit Circle \(-1\) \(1\) \(1\) \(-1\)

The General Solution Framework

Multiple Solutions on a Sine Wave

Principal vs. General Solutions

Principal Value General Solution
What it gives One specific angle from the calculator All angles satisfying the equation
Range Restricted domain of inverse function All real numbers (or context domain)
Example for \(\sin\theta=0.5\) \(\frac{\pi}{6}\) only \(\frac{\pi}{6}+2\pi n\) and \(\frac{5\pi}{6}+2\pi n\)
In context May miss valid answers Filter by the problem's domain

Step-by-Step Approach

1
Isolate the trig expression. Use algebra to get a single trig function on one side: \(\sin(Bx + C) = k\).
2
Apply the inverse function. Find the principal value: \(Bx + C = \sin^{-1}(k)\).
3
Find the second solution in one period. For sine: \(\pi - \theta_1\). For cosine: \(2\pi - \theta_1\).
4
Write the general solution. Add \(+\,2\pi n\) (or \(+\,\pi n\) for tangent) to each base solution.
5
Solve for the original variable (often time \(t\)) and filter by context: keep only solutions in the valid domain.

Example 1: Temperature Model

A city's average temperature is modeled by:

1
Isolate: \(65 = 20\sin\!\left(\frac{\pi}{6}(t-4)\right) + 55\) gives \(\sin\!\left(\frac{\pi}{6}(t-4)\right) = 0.5\).
2
Inverse: \(\frac{\pi}{6}(t-4) = \frac{\pi}{6}\) or \(\frac{\pi}{6}(t-4) = \frac{5\pi}{6}\).
3
Solve for \(t\): From \(\frac{\pi}{6}(t-4)=\frac{\pi}{6}\), we get \(t-4=1\), so \(t=5\) (May). From \(\frac{5\pi}{6}\), we get \(t-4=5\), so \(t=9\) (September).

Example 1: Graph Confirmation

Example 2: Ferris Wheel

A Ferris wheel has radius 25 m and its center is 30 m above the ground. It completes one revolution in 40 seconds. Height model:

1
Isolate: \(42.5 = -25\cos\!\left(\frac{\pi}{20}t\right)+30\) gives \(\cos\!\left(\frac{\pi}{20}t\right)=-0.5\).
2
Inverse: \(\frac{\pi}{20}t = \cos^{-1}(-0.5) = \frac{2\pi}{3}\). Second solution: \(\frac{4\pi}{3}\).
3
Solve: \(t = \frac{20}{\pi}\cdot\frac{2\pi}{3} = \frac{40}{3} \approx\) 13.3 s and \(t = \frac{20}{\pi}\cdot\frac{4\pi}{3} = \frac{80}{3} \approx\) 26.7 s.

Example 2: Visualizing the Ride

Example 3: Tidal Depth

A harbor's water depth (in feet) is:

1
Isolate: \(\sin\!\left(\frac{\pi}{6}t\right) = 0.5\).
2
Inverse: \(\frac{\pi}{6}t = \frac{\pi}{6}\) or \(\frac{5\pi}{6}\).
3
Solve: \(t = 1\) hour (1:00 AM) and \(t = 5\) hours (5:00 AM). Both are in \([0, 12]\). Interpretation: the depth reaches 12 ft while the tide is rising and again while it is falling.

Common Mistakes

Only one solution
Using only the calculator's principal value and forgetting the second solution within each period.
Ignoring context domain
Reporting solutions outside the valid range (e.g., negative time, or month 14).
Wrong period formula
Using \(+\,2\pi n\) for tangent equations instead of the correct \(+\,\pi n\).

Summary: Solution Formulas

Equation Solutions in \([0, 2\pi)\) General Solution
\(\sin\theta = k\) \(\sin^{-1}(k)\) and \(\pi - \sin^{-1}(k)\) \(\theta_1 + 2\pi n,\;\theta_2 + 2\pi n\)
\(\cos\theta = k\) \(\cos^{-1}(k)\) and \(2\pi - \cos^{-1}(k)\) \(\pm\,\theta_1 + 2\pi n\)
\(\tan\theta = k\) \(\tan^{-1}(k)\) and \(\pi + \tan^{-1}(k)\) \(\theta_1 + \pi n\)

Question 1 of 5

The temperature in a city is modeled by \(T(t) = 15\sin\!\left(\frac{\pi}{6}(t-3)\right)+60\), where \(t\) is the month. What is the first month the temperature reaches \(67.5^\circ\text{F}\)?

Question 2 of 5

If \(\cos(\theta) = -0.6\), how many solutions exist in the interval \([0, 2\pi)\)?

Question 3 of 5

A Ferris wheel's height is \(h(t)=-20\cos\!\left(\frac{\pi}{15}t\right)+22\). When does the rider first reach 32 m?

Question 4 of 5

What is the general solution for \(\tan(\theta) = 1\)?

Question 5 of 5

Water depth is \(d(t) = 3\sin\!\left(\frac{\pi}{6}t\right) + 8\). How many times does the depth reach 9.5 ft in a full 24-hour day (\(0 \le t \le 24\))?