# Trigonometric equations

> IB Mathematics: Analysis and Approaches HL · Geometry & Trigonometry
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u3-trigonometric-equations/

This module covers solving linear and quadratic trigonometric equations over restricted and unrestricted domains, using identities and periodicity to find all valid solutions, aligned with IB AA HL assessment expectations.

**Prerequisites:** [Trigonometric identities](https://www.owlsprep.com/study/ib-math-aa-hl-u3-trigonometric-identities/); [Unit circle and trigonometric periodicity](https://www.owlsprep.com/study/ib-math-aa-hl-u3-trigonometric-functions/)

## Learning objectives

- Solve linear trigonometric equations over a given domain
- Solve quadratic trigonometric equations using trigonometric identities
- Find general solutions accounting for periodicity of trigonometric functions
- Avoid common traps related to domain restrictions and lost solutions

## Linear Trigonometric Equations

A linear trigonometric equation has the form $a\sin(bx + c) = k$, $a\cos(bx + c) = k$, or $a\tan(bx + c) = k$, where $a, b, c, k$ are constants. We first isolate the trigonometric term, then find solutions using the unit circle.

**Linear trigonometric equation** — An equation where the trigonometric function is raised only to the first power, with no products of different trigonometric terms.

**Worked example:** Solve $2\sin x = 1$ for $0 \leq x < 2\pi$

1. First, isolate $\sin x$:
2. $$\sin x = \frac{1}{2}$$
3. Find the principal solution in the first quadrant: $x = \frac{\pi}{6}$. Sine is positive in both the first and second quadrants, so the second solution in $[0, 2\pi)$ is:
4. $$x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$
5. Both solutions fall within the given domain, so the final solutions are:
6. $$x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6}$$

> **tip**
>
> Always check the sign of the trigonometric ratio to identify which quadrants your solutions lie in before finalizing your answer.

## Quadratic Trigonometric Equations

Quadratic trigonometric equations can be rearranged into the form $au^2 + bu + c = 0$, where $u$ is a trigonometric function like $\sin x$, $\cos x$, or $\tan x$. We solve using factoring or the quadratic formula, then solve each resulting linear trigonometric equation.

**Worked example:** Solve $2\cos^2 x - 3\cos x + 1 = 0$ for $0 \leq x \leq \pi$

1. Substitute $u = \cos x$ to get a standard quadratic:
2. $$2u^2 - 3u + 1 = 0$$
3. Factor the quadratic:
4. $$(2u - 1)(u - 1) = 0$$
5. This gives two cases: $u = 1$ or $u = \frac{1}{2}$. Solve each case for $x$:
6. Case 1: $\cos x = 1 \implies x = 0$, which is in the domain. Case 2: $\cos x = \frac{1}{2} \implies x = \frac{\pi}{3}$ (the other solution $\frac{5\pi}{3}$ is greater than $\pi$, so excluded).
7. Final solutions:
8. $$x = 0, \quad x = \frac{\pi}{3}$$

## Solving Equations Using Identities

Many trigonometric equations require use of double-angle, compound-angle, or Pythagorean identities to simplify them into a solvable linear or quadratic form. Always look for opportunities to rewrite multiple angles or different trigonometric terms in terms of a single variable.

**Worked example:** Solve $\sin 2x = \sin x$ for $0 \leq x < 2\pi$

1. Use the double-angle identity $\sin 2x = 2\sin x \cos x$:
2. $$2\sin x \cos x = \sin x$$
3. Rearrange and factor out $\sin x$ (do not divide both sides by $\sin x$, this will lose solutions):
4. $$2\sin x \cos x - \sin x = 0 \implies \sin x (2\cos x - 1) = 0$$
5. Set each factor equal to zero and solve: Case 1: $\sin x = 0 \implies x = 0, \pi$. Case 2: $2\cos x - 1 = 0 \implies \cos x = \frac{1}{2} \implies x = \frac{\pi}{3}, \frac{5\pi}{3}$.
6. All solutions are within the domain, so the full set of solutions is:
7. $$x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}$$

## General Solutions for Unrestricted Domains

When no domain is specified, we need to write the general solution that accounts for the periodicity of trigonometric functions. Sine and cosine have period $2\pi$, so we add $2\pi n$ to all base solutions. Tangent has period $\pi$, so we add $\pi n$ to base solutions, where $n \in \mathbb{Z}$.

**Worked example:** Find the general solution of $\tan\left(x - \frac{\pi}{4}\right) = 1$

1. Let $\theta = x - \frac{\pi}{4}$. The general solution of $\tan \theta = 1$ is:
2. $$\theta = \frac{\pi}{4} + \pi n, \quad n \in \mathbb{Z}$$
3. Substitute back $\theta = x - \frac{\pi}{4}$:
4. $$x - \frac{\pi}{4} = \frac{\pi}{4} + \pi n$$
5. Rearrange to get the general solution:
6. $$x = \frac{\pi}{2} + \pi n, \quad n \in \mathbb{Z}$$

> **warning**
>
> Tangent has period $\pi$, not $2\pi$! Adding $2\pi n$ for tangent will leave out half of all valid solutions.

## Common pitfalls

- **Wrong:** Only taking the principal calculator solution and ignoring solutions in other quadrants
  - Why it fails: Calculators only output one principal solution, but trigonometric equations have multiple solutions per period
  - Correct: Use the unit circle and the sign of the trigonometric ratio to find all base solutions in one period before adding period multiples
- **Wrong:** Dividing both sides of an equation by a common trigonometric term (e.g. dividing $\sin 2x = \sin x$ by $\sin x$)
  - Why it fails: This removes the case where the trigonometric term equals zero, losing all corresponding solutions
  - Correct: Rearrange the equation to factor out the common trigonometric term, keeping all possible solution cases
- **Wrong:** Adding $2\pi n$ as the period for tangent equations
  - Why it fails: Tangent repeats every $\pi$, not $2\pi$, so this misses half of all general solutions
  - Correct: Always add $\pi n$ for tangent equations, and $2\pi n$ for sine and cosine equations
- **Wrong:** Failing to filter solutions to match the stated domain
  - Why it fails: After generating general solutions, it is easy to include solutions that fall outside the given interval
  - Correct: List all candidate solutions, then explicitly check each one to confirm it falls within the domain bounds before writing your final answer
- **Wrong:** Leaving extraneous solutions untested after squaring both sides of an equation
  - Why it fails: Squaring both sides can introduce solutions that do not satisfy the original equation
  - Correct: Test every solution in the original equation to remove any invalid extraneous solutions

## Cheatsheet

| Equation Type | Key Method | General Solution Form |
| --- | --- | --- |
| $\sin \theta = k$ | Find base solutions via symmetry | $\theta = \alpha + 2\pi n, \theta = \pi - \alpha + 2\pi n, n \in \mathbb{Z}$ |
| $\cos \theta = k$ | Find base solutions via symmetry | $\theta = \alpha + 2\pi n, \theta = -\alpha + 2\pi n, n \in \mathbb{Z}$ |
| $\tan \theta = k$ | One base solution per period | $\theta = \alpha + \pi n, n \in \mathbb{Z}$ |
| Quadratic trig | Substitute $u = \sin x/\cos x/\tan x$ | Factor/solve quadratic, solve each linear case |
| Equation with multiple angles | Use angle identities to simplify | Adjust period for the multiple angle coefficient |

## What's next

Solving trigonometric equations is a core foundational skill for almost all remaining topics in IB AA HL. It is used regularly when integrating trigonometric functions, solving differential equations, analysing periodic models, and working with polar forms of complex numbers. Mastery of this topic will make working with these more advanced concepts significantly easier, as you will already be comfortable rearranging and solving trigonometric expressions. Next, you can deepen your understanding of trigonometric applications or extend your knowledge to related topics.

- [Compound and double angle identities](https://www.owlsprep.com/study/ib-math-aa-hl-u3-compound-and-double-angle-identities/)
- [2D Coordinate Geometry](https://www.owlsprep.com/study/ib-math-aa-hl-u3-2d-coordinate-geometry/)
- [3D coordinate geometry](https://www.owlsprep.com/study/ib-math-aa-hl-u3-3d-coordinate-geometry/)

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