Trig Equations with Identities and Factoring
An equation like $2\sin^2 x - \sin x = 0$ has no single inverse-trig button that solves it, because it mixes a squared term with a linear one. The move that works is the same one that cracks a polynomial: get everything on one side, factor, and set each factor to zero.
By the end of this page you can:
- Recognize when a trig equation must be rewritten with an identity before it can be solved.
- Factor a trig equation and solve each factor as a basic trig equation.
- List every solution in a given interval without dropping any.
The floor. Move all terms to one side, factor, and solve each factor separately. Never divide both sides by a trig expression, because that throws away the solutions where the expression equals zero.
Quick Reference
| Field | Value |
|---|---|
| Textbook | Stewart 7e |
| Chapter | 7 |
| Section | 7.5 More Trigonometric Equations |
| Open alternate | OpenStax Precalculus 2e, 7.6 Modeling with Trigonometric Equations |
| Difficulty | Core |
| Time | 30 minutes |
Before You Start
The Idea
A factorable trig equation is one where every term shares a common factor, or where a substitution turns it into a quadratic in one trig function. The strategy: collect all terms on one side so the other side is zero, factor, then use the zero-product property. Each factor set to zero is a basic trig equation you already know how to solve.
Two warnings carry most of the mistakes. First, do not divide out a common trig factor. Dividing $2\sin^2 x = \sin x$ by $\sin x$ silently deletes the solutions where $\sin x = 0$. Factoring keeps them. Second, when a problem hides a squared term behind an identity, such as $2\cos^2 x + \sin x = 1$, rewrite $\cos^2 x$ as $1 - \sin^2 x$ first so the whole equation is in one function, then factor.
Key Formulas
The zero-product property is the engine:
\[ A \cdot B = 0 \quad \Longleftrightarrow \quad A = 0 \ \text{ or } \ B = 0. \]
The identity most often needed to make an equation factorable:
\[ \sin^2 x + \cos^2 x = 1. \]
Source: Stewart 7e, Section 7.5 More Trigonometric Equations.
Worked Example
Solve $2\sin^2 x - \sin x = 0$ on $[0, 2\pi)$.
Step 1. The right side is already zero, so factor the left side. Both terms share $\sin x$:
\[ \sin x \,(2\sin x - 1) = 0. \]
Step 2. Set each factor to zero:
\[ \sin x = 0 \qquad \text{or} \qquad 2\sin x - 1 = 0. \]
Step 3. Solve each basic equation on $[0, 2\pi)$. From $\sin x = 0$: $x = 0$ and $x = \pi$. From $2\sin x - 1 = 0$, that is $\sin x = \tfrac{1}{2}$: $x = \tfrac{\pi}{6}$ and $x = \tfrac{5\pi}{6}$.
Show the full solution set
\[ x = 0, \ \tfrac{\pi}{6}, \ \tfrac{5\pi}{6}, \ \pi. \]
All four values lie in $[0, 2\pi)$. Note that dividing the original equation by $\sin x$ would have lost $x = 0$ and $x = \pi$, which is exactly why factoring is the safe route.
Connections
This builds on solving basic trig equations, on simplifying and proving trig identities, and on the double-angle formulas, since a $\sin 2x$ or $\cos 2x$ term is usually replaced before factoring. It is the bridge from single-function equations to the messier equations that appear in modeling periodic motion, where you set a sinusoidal model equal to a target value and solve for the times it occurs.
Resources
- Stewart 7e, Section 7.5 More Trigonometric Equations (primary).
- OpenStax Precalculus 2e, Section 7.6 Modeling with Trigonometric Equations (alternate).
Last updated: 2026-06-24