Solving trigonometric equations

IB Mathematics: Analysis and Approaches SL· 30 min read

1. Linear Trigonometric Equations over Restricted Domains★★☆☆☆⏱ 15 min

📘 Definition

Linear trigonometric equation

$a sin (b x + c) + d = k$

An equation containing only one trigonometric term, raised to the first power, that can be rearranged to isolate the trigonometric term on one side.

Example:

$2 sin 3 x = 1$

For IB AA SL, you will most often be given a restricted domain, a closed interval for $x$ , and asked to find all solutions that lie within this interval. The core process is to isolate the trigonometric term first, then find all matching angles using the symmetry and periodicity of the function.

📐 Worked Example

Solve $2 sin x = 1$ for $0 \leq x < 2 π$

1
Isolate the trigonometric term by dividing both sides by 2
2
$\sin x = \frac{1}{2}$
3
Find the principal solution using the unit circle or inverse sine
4
$x = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$
5
Sine is positive in the first and second quadrants, so find the second solution in $[0, 2 π)$
6
$x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
7
Both solutions fall within the given domain, so these are the final results

Exam tip:

Always check that every solution you list falls inside the given domain; extra solutions outside the domain will cost you marks.

2. General Solutions of Trigonometric Equations★★★☆☆⏱ 20 min

When asked to find all solutions over $x \in R$ , you need to account for the repeating nature of trigonometric functions by adding a multiple of the function's period to your base solutions.

📘 Definition

General solution

The complete set of all possible solutions, written by adding an integer multiple of the period to each base solution over one full period of the function.

📐 Worked Example

Find the general solution of $2 sin x = 1$

1
Find the base solutions over one full period of $sin x$ (period = $2 π$ )
2
$x_1 = \frac{\pi}{6}, \quad x_2 = \frac{5\pi}{6}$
3
Add an integer multiple of the period $2 π$ to each base solution, where $k \in Z$
4
$x = \frac{\pi}{6} + 2k\pi, \quad x = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$

3. Quadratic Trigonometric Equations★★★☆☆⏱ 20 min

Quadratic trigonometric equations have a squared trigonometric term, and can be solved with the same method as standard quadratics: substitute to get a quadratic in one variable, factor or use the quadratic formula, then solve each resulting linear trig equation separately.

📐 Worked Example

Solve $2 cos^{2} x - 3 cos x + 1 = 0$ for $0 \leq x < 2 π$

1
Substitute $u = cos x$ to rewrite the equation as a standard quadratic
2
$2u^2 - 3u + 1 = 0$
3
Factor the quadratic and solve for $u$
4
$(2u - 1)(u - 1) = 0 \implies u = \frac{1}{2} \text{ or } u = 1$
5
Solve $cos x = 1$ over the given domain
6
$x = 0$
7
Solve $cos x = \frac{1}{2}$ : cosine is positive in Q1 and Q4
8
$x = \frac{\pi}{3}, \quad x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$
9
Collect all solutions that fall within the domain: $0, \frac{π}{3}, \frac{5 π}{3}$

4. Equations Requiring Identity Rearrangement★★★★☆⏱ 25 min

🚫 No Calculator

Many exam questions give equations with mixed trigonometric terms that require you to use identities (most often the Pythagorean identity) to rearrange into a solvable linear or quadratic form.

📐 Worked Example

Solve $2 sin^{2} x = cos x + 1$ for $0 \leq x < 2 π$

1
Use the Pythagorean identity $sin^{2} x = 1 - cos^{2} x$ to rewrite the equation in terms of $cos x$ only
2
$2(1 - \cos^2 x) = \cos x + 1$
3
Expand and rearrange into standard quadratic form
4
$2\cos^2 x + \cos x - 1 = 0$
5
Factor the quadratic
6
$(2\cos x - 1)(\cos x + 1) = 0 \implies \cos x = \frac{1}{2} \text{ or } \cos x = -1$
7
Solve each linear trig equation over the domain
8
$\cos x = \frac{1}{2} \implies x = \frac{\pi}{3}, \frac{5\pi}{3}; \quad \cos x = -1 \implies x = \pi$
9
All solutions are within the domain, so final solutions are $\frac{π}{3}, π, \frac{5 π}{3}$

5. Common Pitfalls

Wrong move:

Forgetting to add the period to every base solution when finding the general solution

Why:

Only adding the period to one solution leaves out half of the valid solutions for sine and cosine equations

Correct move:

Add the integer multiple of the period to every distinct base solution found over one full period

Wrong move:

Canceling a common trigonometric term from both sides of an equation instead of factoring

Why:

Canceling removes all solutions where the trigonometric term equals zero, leading to missing solutions

Correct move:

Move all terms to one side of the equation and factor out the common trigonometric term

Wrong move:

Including solutions that fall outside the given restricted domain

Why:

Adding or subtracting periods to base solutions creates solutions outside the required interval

Correct move:

List all candidate solutions, then filter out any that do not satisfy the domain inequality

Wrong move:

Using the wrong period for scaled tangent functions like $tan 2 x$

Why:

Tangent has a period of $π$ , not $2 π$ , so scaled tangent has half the period you might expect

Correct move:

For $tan (b x)$ , always use period $\frac{π}{∣ b ∣}$

Wrong move:

Keeping solutions where $∣ sin x ∣ > 1$ or $∣ cos x ∣ > 1$

Why:

These values are outside the range of sine and cosine, so they have no corresponding real solutions

Correct move:

Discard any such roots immediately after solving for the trigonometric term

6. Quick Reference Cheatsheet

Equation Type	Solution Process
Linear	Isolate the trigonometric term 2. Find all base solutions over one period 3. Add $k \times period$ for general solutions, filter for restricted domain
Quadratic	Substitute $u = sin x / cos x / tan x$ 2. Solve quadratic for $u$ 3. Discard values outside function range 4. Solve each resulting linear equation
Mixed (needs identity)	Use Pythagorean/other identity to get one trig variable 2. Rearrange to linear/quadratic 3. Solve as above 4. Check for extraneous solutions
Period Rule	$sin (b x), cos (b x) : T = \frac{2 π}{∣ b ∣}$ ; $tan (b x) : T = \frac{π}{∣ b ∣}$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · Paper 1
Solve sin x = 0.5 over 0 ≤ x < 2π
2022 · Paper 2
Quadratic in cos x over -π ≤ x < π
2023 · Paper 1
Find general solution of tan 2x = 1

Going deeper

syllabusIB AA SL Official Syllabus
practiceIB Past Trig Equation QuestionsAvailable via your school's IB question bank

What's Next

Solving trigonometric equations is a core foundational skill that appears across all units of IB AA SL. You will use it to find critical points in calculus, calculate angles between vectors, and solve for unknowns in periodic modeling problems. It is heavily tested in both Paper 1 and Paper 2 of your final exam, so practicing finding all solutions is critical for good marks. Next, you will apply this skill to triangle problems with the sine and cosine rule, then build on it to model real-world periodic phenomena with trigonometric functions.