Solving Basic Trig Equations
A Ferris wheel rider sits at a height the manufacturer models with a sine function. To find every clock time when she is exactly halfway up, you have to solve an equation like sin x = 1/2, and you need every answer, not just the one a calculator hands back. A basic trig equation sets a single trig function equal to a number, and because the function repeats, it has infinitely many solutions.
By the end of this page you can:
- Find one solution of sin x = c, cos x = c, or tan x = c using an inverse trig value.
- Use the reference angle and the sign to find the second solution inside one cycle.
- Write every solution by adding the period to each one.
The floor. A basic trig equation usually has two solutions in one full cycle, and you get all solutions by adding the period (2 pi for sine and cosine) to each of those.
Quick Reference
| Field | Value |
|---|---|
| Textbook | Stewart, Redlin, Watson 7e |
| Chapter | 7 |
| Section | 7.4 Basic Trigonometric Equations |
| Open alternate | OpenStax Precalculus 2e, 7.5 Solving Trigonometric Equations |
| Difficulty | core |
| Time | about 30 minutes |
Before You Start
The Idea
Solving a basic trig equation runs in two stages. First find one angle whose function value matches the target. The inverse function gives this first angle directly. Second, use the fact that sine and cosine each take the same value at two angles in one trip around the circle, so the reference angle and the sign of the target locate the second angle. Once you have both angles inside one cycle, every other solution is one of those two angles shifted by a whole number of full turns.
For sine and cosine the full turn is 2 pi, so you append 2 pi n, where n is any integer. For tangent the period is pi, so its solutions repeat every pi instead.
Key Formulas
For sine, with one solution x = a in [0, 2 pi), the second solution in that cycle is pi minus a, and all solutions are:
\[ x = a + 2\pi n \quad \text{and} \quad x = (\pi - a) + 2\pi n, \quad n \in \mathbb{Z} \]
For cosine, with one solution x = a, the second is its negative, and all solutions are:
\[ x = a + 2\pi n \quad \text{and} \quad x = -a + 2\pi n, \quad n \in \mathbb{Z} \]
For tangent, with one solution x = a, all solutions are:
\[ x = a + \pi n, \quad n \in \mathbb{Z} \]
Source: Stewart 7e, Section 7.4 Basic Trigonometric Equations.
Worked Example
Solve sin x = 1/2 for all real x.
Step 1. Find one solution. The inverse sine of 1/2 is pi/6, so x = pi/6 is a solution.
Step 2. Find the second solution in one cycle. Sine is positive in the first and second quadrants, so the other angle has the same reference angle pi/6 but sits in the second quadrant. That angle is pi minus pi/6, which is 5 pi/6. Check: sin(5 pi/6) = 1/2.
Step 3. Add the period. Sine repeats every 2 pi, so shift each solution by 2 pi n.
Show all solutions
\[ x = \frac{\pi}{6} + 2\pi n \quad \text{and} \quad x = \frac{5\pi}{6} + 2\pi n, \quad n \in \mathbb{Z} \]
The two base angles pi/6 and 5 pi/6 are the only ones in [0, 2 pi); every other solution is one of them plus a whole number of full turns.
Connections
This builds on inverse trig functions, which produce the first solution, on reference angles, which place the second solution in the right quadrant, and on the unit circle for sine and cosine, which shows why a value repeats twice per cycle. It leads next to solving trig equations that first need an identity or factoring before they reduce to a basic equation of this form.
Resources
- Stewart, Redlin, Watson, Precalculus 7e, Section 7.4 Basic Trigonometric Equations (primary).
- OpenStax Precalculus 2e, Section 7.5 Solving Trigonometric Equations (alternate).
Last updated: 2026-06-24