# Solving Trigonometric Equations

> CIE IGCSE Additional Mathematics · 2025-2027
> Source: https://www.owlsprep.com/study/cie-0606-u10-solving-trigonometric-equations/

This guide teaches you to solve trigonometric equations for all six trigonometric functions using core 0606 identities, find all valid solutions in a specified domain, and avoid common exam errors.

**Prerequisites:** Knowledge of the six trigonometric functions, their graphs and symmetry rules; Mastery of core trigonometric reciprocal and Pythagorean identities

## Learning objectives

- Solve linear trigonometric equations for all six trigonometric functions over a specified domain
- Use core trigonometric identities to convert equations into solvable linear or quadratic forms
- Identify all valid solutions within a given interval, avoiding missing or extraneous solutions
- Apply non-calculator and calculator solving techniques appropriate for Papers 1 and 2 of the 0606 exam

## Linear Trigonometric Equations Over a Specified Domain

Linear trigonometric equations have the form $k \times \text{trig}(\theta) = c$, where $k$ and $c$ are constants, and $\text{trig}(\theta)$ is any of the six trigonometric functions. To solve these, first isolate the trigonometric function, then find all solutions within the given domain using symmetry and periodicity rules.

**Principal Solution** — The first (least non-negative) solution of a trigonometric equation, lying in $[0^\circ, 360^\circ)$ for degrees or $[0, 2\pi)$ for radians. This is not the same as a calculator's 'principal value' (the inverse-function output range).

**Worked example:** Solve $2\sin\theta + 1 = 0$ for $\theta$ in the interval $0^\circ \leq \theta < 360^\circ$.

1. Isolate the trigonometric function:

   $$2\sin\theta = -1 \implies \sin\theta = -\frac{1}{2}$$
2. Find the principal acute solution for the positive value:

   $$\sin^{-1}\left(\frac{1}{2}\right) = 30^\circ$$
3. Sine is negative in the 3rd and 4th quadrants, so calculate solutions:

   $$180^\circ + 30^\circ = 210^\circ, 360^\circ - 30^\circ = 330^\circ$$
4. Verify both solutions lie within the given interval. Final solutions: $\theta = 210^\circ, 330^\circ$

> **Exam tip:** For Paper 1 non-calculator questions, all solutions will be multiples of 30°, 45°, 60° or 90°, so memorize exact trigonometric values for these angles.

## Solving Equations Using Reciprocal Identities

Reciprocal identities (e.g. $\sec\theta = \frac{1}{\cos\theta}$, $\cot\theta = \frac{1}{\tan\theta}$) let you rewrite equations involving sec, cosec, or cot into equations using sin, cos, tan, which are easier to solve. Always remember to check for extraneous solutions where the original function is undefined.

> **warning**
>
> Never multiply or divide both sides of an equation by a trigonometric function without first noting the values where that function is zero or undefined, as this can introduce or remove solutions.

**Worked example:** Solve $4\cot\theta = \tan\theta$ for $0 \leq \theta < 2\pi$.

1. Rewrite cot using reciprocal identity, noting $\tan\theta \neq 0$:

   $$4 \times \frac{1}{\tan\theta} = \tan\theta$$
2. Rearrange to eliminate the denominator:

   $$4 = \tan^2\theta \implies \tan\theta = \pm 2$$
3. Find the principal solution for the positive value:

   $$\tan^{-1}(2) \approx 1.107 \text{ radians}$$
4. Tan is positive in 1st and 3rd quadrants, negative in 2nd and 4th. Calculate all solutions: $1.107$, $1.107 + \pi \approx 4.249$, $\pi - 1.107 \approx 2.034$, $2\pi - 1.107 \approx 5.176$
5. Verify none of these make $\tan\theta = 0$, so all are valid. Final solutions (3 s.f.): $1.11, 2.03, 4.25, 5.18$ radians

*Calculator:* allowed

## Solving Quadratic Trigonometric Equations Using Pythagorean Identities

Quadratic trigonometric equations involve squared terms of trigonometric functions. Use Pythagorean identities (e.g. $\sec^2\theta = 1 + \tan^2\theta$) to rewrite the equation as a quadratic in a single trigonometric function, then factor or use the quadratic formula to solve.

**Extraneous Solution** — A solution obtained during algebraic manipulation that does not satisfy the original equation, usually introduced by squaring both sides or using identities that restrict the domain.

**Worked example:** Solve $2\sec^2\theta + \tan\theta - 3 = 0$ for $0^\circ \leq \theta < 360^\circ$.

1. Replace $\sec^2\theta$ using the Pythagorean identity:

   $$2(1 + \tan^2\theta) + \tan\theta - 3 = 0$$
2. Simplify to standard quadratic form:

   $$2\tan^2\theta + \tan\theta - 1 = 0$$
3. Factor the quadratic expression:

   $$(2\tan\theta - 1)(\tan\theta + 1) = 0$$
4. Solve first factor $2\tan\theta -1 = 0$: $\tan\theta = 0.5$. Solutions: $26.6^\circ$, $206.6^\circ$
5. Solve second factor $\tan\theta + 1 =0$: $\tan\theta = -1$. Solutions: $135^\circ$, $315^\circ$
6. Verify all solutions are valid (sec is defined for all values here). Final solutions: $26.6^\circ, 135^\circ, 206.6^\circ, 315^\circ$

> **Exam tip:** If you use the quadratic formula to solve for a trigonometric function, discard any values outside the range of that function (e.g. values of $\sin\theta$ greater than 1 or less than -1).

## Solving $a\sin\theta + b\cos\theta = 0$ Equations

For equations of the form $a\sin\theta + b\cos\theta = 0$, you can rearrange and divide by $\cos\theta$ (as long as $\cos\theta \neq 0$) to get a tan equation, which you can solve using standard methods. The R-formula for $a\sin\theta + b\cos\theta = c$ (where $c \neq 0$) is A-Level content and out of scope for 0606, so these equations will not appear on your exam.

**Worked example:** Solve $3\sin\theta - 2\cos\theta =0$ for $0^\circ \leq \theta < 360^\circ$.

1. Rearrange the equation to group terms:

   $$3\sin\theta = 2\cos\theta$$
2. Divide both sides by $\cos\theta$, noting $\cos\theta \neq 0$ (no solutions lost as $\cos\theta=0$ would make LHS=±3, RHS=0):

   $$3\tan\theta = 2 \implies \tan\theta = \frac{2}{3} \approx 0.6667$$
3. Find solutions: $\tan^{-1}(2/3) \approx 33.7^\circ$, $33.7^\circ + 180^\circ = 213.7^\circ$
4. Verify both solutions are valid. Final solutions: $33.7^\circ, 213.7^\circ$

## Solving Equations with Multiple or Shifted Angles

Many exam equations act on a multiple or shifted angle, such as $\sin 2x$, $\cos 3\theta$ or $\tan(\theta - 30^\circ)$. To solve $\text{trig}(bx + d) = k$, substitute $u = bx + d$, transform the given domain for $x$ into the matching domain for $u$, solve for every value of $u$ in that expanded domain, then reverse the substitution with $x = \frac{u - d}{b}$ to recover each value of $x$.

> **warning**
>
> The expanded $u$-domain is the whole point of this method. Because $u = bx + d$ runs $b$ times faster than $x$, the interval for $u$ is $b$ times longer, so there are about $b$ times as many solutions. Solving only within the original interval and forgetting the extra periods is the most common way to lose marks here.

**Worked example:** Solve $\sin 2x = \frac{\sqrt{3}}{2}$ for $0^\circ \leq x < 360^\circ$.

1. Substitute $u = 2x$ and transform the domain. Multiplying $0^\circ \leq x < 360^\circ$ through by 2 gives:

   $$0^\circ \leq u < 720^\circ$$
2. Solve $\sin u = \frac{\sqrt{3}}{2}$. Reference angle $\alpha = 60^\circ$; sine is positive in the 1st and 2nd quadrants, giving the first-revolution solutions:

   $$u = 60^\circ,\ 120^\circ$$
3. Add $360^\circ$ to reach the second revolution, keeping only values below $720^\circ$:

   $$u = 420^\circ,\ 480^\circ$$
4. Reverse the substitution with $x = \frac{u}{2}$ for all four values of $u$:

   $$x = 30^\circ,\ 60^\circ,\ 210^\circ,\ 240^\circ$$

**Worked example:** Solve $\sin\left(2\theta + \frac{\pi}{6}\right) = \frac{1}{2}$ for $0 \leq \theta < 2\pi$. Give exact answers in terms of $\pi$.

1. Substitute $u = 2\theta + \frac{\pi}{6}$ and transform the domain. As $\theta$ runs over $0 \leq \theta < 2\pi$, $u$ runs from $\frac{\pi}{6}$ up to (but not including) $2(2\pi) + \frac{\pi}{6} = \frac{25\pi}{6}$:

   $$\frac{\pi}{6} \leq u < \frac{25\pi}{6}$$
2. Solve $\sin u = \frac{1}{2}$. Reference angle $\alpha = \frac{\pi}{6}$; sine is positive in the 1st and 2nd quadrants, giving base solutions (both lie in the domain, which starts at $\frac{\pi}{6}$):

   $$u = \frac{\pi}{6},\ \frac{5\pi}{6}$$
3. Add $2\pi$ for the next period; the value $\frac{25\pi}{6}$ is the excluded upper limit, so stop here:

   $$u = \frac{13\pi}{6},\ \frac{17\pi}{6}$$
4. Reverse the substitution with $\theta = \frac{u - \frac{\pi}{6}}{2}$ for each value of $u$:

   $$\theta = 0,\ \frac{\pi}{3},\ \pi,\ \frac{4\pi}{3}$$

> **Exam tip:** Transform the domain immediately after substituting, before you solve. Write the new $u$-interval at the top of your working so you remember to collect the extra solutions from each additional period.

## Common pitfalls

- **Wrong:** Only giving the principal solution and forgetting other solutions in the domain
  - Why it fails: Trigonometric functions are periodic, so there are often multiple solutions in a given interval
  - Correct: After finding the principal solution, use quadrant symmetry and periodicity rules to find all solutions that fit the stated domain, then check each one
- **Wrong:** Dividing both sides of an equation by a trigonometric function without checking for solutions where that function is zero
  - Why it fails: This removes valid solutions where the function equals zero
  - Correct: First rearrange the equation to factor out the common trigonometric function, then solve each factor separately, or note excluded values and check them separately after solving
- **Wrong:** Using the R-formula to solve $a\sin\theta + b\cos\theta = c$ for $c \neq 0$
  - Why it fails: The R-formula is A-Level content, out of scope for 0606, and these equations will not appear on the exam
  - Correct: Only solve $a\sin\theta + b\cos\theta = 0$ by rearranging to get a tan equation as specified in the syllabus
- **Wrong:** Forgetting to check for extraneous solutions after squaring both sides or using identities
  - Why it fails: Squaring both sides can introduce solutions that do not satisfy the original equation
  - Correct: Substitute every solution you find back into the original equation to confirm it is valid before including it in your final answer
- **Wrong:** Mixing up degrees and radians when solving equations
  - Why it fails: The question specifies the unit in the domain, and using the wrong unit leads to incorrect answers, especially on calculator papers
  - Correct: Always set your calculator to the unit specified in the domain (degrees or radians) before solving, and confirm your solutions match the unit of the given interval

## Cheatsheet

| Trigonometric Function | Symmetry Rule for Solutions (0° ≤ θ < 360°) | Period |
| --- | --- | --- |
| $\sin\theta = k$ | If k positive: $\alpha, 180^\circ - \alpha$; If k negative: $180^\circ + \alpha, 360^\circ - \alpha$ where $\alpha = \sin^{-1}(\|k\|)$ | 360° / 2π |
| $\cos\theta = k$ | If k positive: $\alpha, 360^\circ - \alpha$; If k negative: $180^\circ - \alpha, 180^\circ + \alpha$ where $\alpha = \cos^{-1}(\|k\|)$ | 360° / 2π |
| $\tan\theta = k$ | If k positive: $\alpha, 180^\circ + \alpha$; If k negative: $180^\circ - \alpha, 360^\circ - \alpha$ where $\alpha = \tan^{-1}(\|k\|)$ | 180° / π |
| Reciprocal Identities | $\cosec\theta = 1/\sin\theta$, $\sec\theta = 1/\cos\theta$, $\cot\theta = 1/\tan\theta$ |  |
| Pythagorean Identities | $\sin^2\theta + \cos^2\theta = 1$, $1 + \tan^2\theta = \sec^2\theta$, $1 + \cot^2\theta = \cosec^2\theta$ |  |

## What's next

Now that you have mastered solving trigonometric equations, you can apply this skill to more advanced trigonometry problems, including calculus questions involving differentiation and integration of trigonometric functions, as well as vector problems with angle calculations. This topic is heavily tested on both Paper 1 and Paper 2 of the CIE IGCSE Additional Mathematics 0606 exam, so practice past paper questions to build speed and accuracy, and make sure you can solve both non-calculator and calculator questions efficiently.

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/cie-0606-u10-solving-trigonometric-equations/