What are trigonometric equations and how are they used in the Cambridge IGCSE Mathematics curriculum?

Trigonometric equations are mathematical expressions that involve trigonometric functions like sine, cosine, and tangent. In the Cambridge IGCSE Mathematics curriculum, students learn to solve these equations to find angles or lengths in various geometric contexts, such as triangles and circles. Understanding these equations is crucial for solving problems related to wave patterns, oscillations, and other real-world applications.

How do you solve basic trigonometric equations involving sine and cosine?

To solve basic trigonometric equations involving sine and cosine, you first isolate the trigonometric function. For example, in an equation like sin(x) = 0.5, you find the angles x that satisfy this condition. Using the unit circle or trigonometric tables, you determine the principal value and then find additional solutions by considering the periodic nature of sine and cosine functions. For sine, solutions are of the form x = arcsin(0.5) + 2πk or x = π - arcsin(0.5) + 2πk, where k is an integer.

What strategies can be used to solve more complex trigonometric equations?

For more complex trigonometric equations, strategies include using trigonometric identities to simplify the equation, such as the Pythagorean identity or angle sum identities. Factoring, using substitution, or converting all terms to a single trigonometric function can also help. For example, in equations involving both sine and cosine, converting everything to sine or cosine using identities can simplify the problem. Graphical methods or numerical solutions may also be employed for equations that are difficult to solve algebraically.

Why is it important to understand the general solutions of trigonometric equations in IGCSE Mathematics?

Understanding the general solutions of trigonometric equations is important in IGCSE Mathematics because it allows students to find all possible solutions within a given domain. Trigonometric functions are periodic, meaning they repeat values at regular intervals. Knowing how to express solutions in terms of a general formula, such as x = nπ + (-1)^n * arcsin(a) for sine, helps students solve equations over different intervals and apply these solutions to real-world problems.

What common mistakes should students avoid when solving trigonometric equations?

Common mistakes include not considering all possible solutions within the given domain, forgetting the periodic nature of trigonometric functions, and misapplying trigonometric identities. Students should also be careful with the use of inverse trigonometric functions, ensuring they account for all angles that satisfy the equation. Additionally, it's important to check solutions in the original equation to avoid extraneous solutions that may arise from algebraic manipulation.

Why is "Giving only the principal value" a common mistake in Trigonometric Equations?

Why it happens: Calculator returns one angle; student stops. How to avoid it: After the principal, USE THE SYMMETRY rule to get the second value in range. Source: 0580/42 — every series

Why is "Including solutions outside the given range" a common mistake in Trigonometric Equations?

Why it happens: Forgetting to check the bounds. How to avoid it: Reject any solution below 0 or above 360 (or whatever range was specified).

Why is "Calculator in radian mode" a common mistake in Trigonometric Equations?

Why it happens: Default mode varies. How to avoid it: Switch to DEG. Sanity check: ^-1(0.5) = 30, not 0.524.

Why is "Solving $\sin 2\theta = k$ as if it were $\sin\theta = k$" a common mistake in Trigonometric Equations?

Why it happens: Treating the multiplied argument as the variable. How to avoid it: Substitute u = 2, find u in the doubled range, then divide by 2 at the end.

TrigonometryCambridge IGCSE Maths 0580 Extended

Trigonometric Equations

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Trigonometric Equations — Cambridge IGCSE 0580 Maths Extended (2026)

Solve $\sin x = k$ , $\cos x = k$ , $\tan x = k$ in a given range. Cambridge expects you to find ALL solutions in $0\degree$ to $360\degree$ — not just the principal value.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E8.6 — Solve simple trigonometric equations for values between $0\degree$ and $360\degree$ .

Solving $\sin x = k$

Principal value $\theta = \sin^{-1}(k)$ . Second solution: $180\degree - \theta$ .

Method.

Compute $\theta = \sin^{-1}(k)$ — the principal value (calculator output).
Second solution in $0$ - $360$ : $180\degree - \theta$ .
Check both solutions lie in the given range.

Worked. Solve $\sin x = 0.6$ for $0\degree \le x \le 360\degree$ .

$\theta = \sin^{-1}(0.6) \approx 36.87\degree$ .
Second: $180\degree - 36.87\degree = 143.13\degree$ .
Both in range: $x \approx 36.9\degree, 143.1\degree$ .

A positive sine value gives two solutions, symmetric about 90°: θ and 180° − θ.

Worked (negative $k$ ). Solve $\sin x = -0.5$ for $0\degree \le x \le 360\degree$ .

Calculator gives $\sin^{-1}(-0.5) = -30\degree$ (out of range).
Use the symmetry: $\sin x < 0$ in the third and fourth quadrants ( $180\degree$ to $360\degree$ ).
Solutions: $180\degree + 30\degree = 210\degree$ and $360\degree - 30\degree = 330\degree$ .

$\sin x = k$ : principal $\theta$ , second $180\degree - \theta$ .
Negative $k$ : solutions in $(180, 360)$ — use $180 + |\theta|$ and $360 - |\theta|$ .
Always check the range.

Solving $\cos x = k$

Principal $\theta = \cos^{-1}(k)$ . Second solution: $360\degree - \theta$ .

Method.

Principal $\theta = \cos^{-1}(k)$ .
Second solution: $360\degree - \theta$ .
Check the range.

Worked. Solve $\cos x = 0.7$ for $0\degree \le x \le 360\degree$ .

$\theta = \cos^{-1}(0.7) \approx 45.57\degree$ .
Second: $360\degree - 45.57\degree = 314.43\degree$ .
$x \approx 45.6\degree, 314.4\degree$ .

Worked (negative $k$ ). Solve $\cos x = -0.4$ for $0\degree \le x \le 360\degree$ .

$\cos^{-1}(-0.4) \approx 113.6\degree$ (calculator gives this directly because $\cos^{-1}$ output is in $0$ - $180$ ).
Second: $360\degree - 113.6\degree = 246.4\degree$ .
$x \approx 113.6\degree, 246.4\degree$ .

$\cos x = k$ : principal $\theta$ , second $360\degree - \theta$ .
Negative $k$ : $\cos^{-1}$ gives an angle in $(90, 180)$ .
Cosine is symmetric across the $y$ -axis: $\cos(360 - x) = \cos x$ .

Solving $\tan x = k$

Principal $\theta = \tan^{-1}(k)$ . Second solution: $\theta + 180\degree$ .

Method.

Principal $\theta = \tan^{-1}(k)$ — gives an angle in $(-90, 90)$ .
Tan period is $180\degree$ , so add $180\degree$ for each subsequent solution.
Adjust to fit $0$ - $360$ .

Worked. Solve $\tan x = 1.5$ for $0\degree \le x \le 360\degree$ .

$\theta = \tan^{-1}(1.5) \approx 56.31\degree$ .
Add $180\degree$ : $236.31\degree$ .
Both in range: $x \approx 56.3\degree, 236.3\degree$ .

Worked (negative $k$ ). Solve $\tan x = -1$ for $0\degree \le x \le 360\degree$ .

$\tan^{-1}(-1) = -45\degree$ (out of range).
Add $180\degree$ : $135\degree$ .
Add another $180\degree$ : $315\degree$ .
$x = 135\degree, 315\degree$ .

$\tan x = k$ : principal $\theta$ , then $\theta + 180\degree$ .
Period is $180\degree$ , not $360\degree$ .
Negative $k$ : principal is negative, add $180\degree$ to bring it in range.

Graphical method — using the trig graph

Sketch the graph. Draw $y = k$ . Read off the $x$ -values where they intersect.

For trickier equations, sketching the graph can help visualise the solutions.

Steps.

Sketch $y = \sin x$ (or $\cos x$ / $\tan x$ ) for the given range.
Draw the horizontal line $y = k$ .
The intersections give the solutions.
Use symmetry to read off both $x$ -values.

ASTC tells you which quadrants give a positive value — read it anticlockwise from quadrant 1.

Worked. Solve $\sin x = -0.7$ for $0\degree \le x \le 360\degree$ .

Sketch sine wave from $0$ to $360\degree$ .
Horizontal line at $y = -0.7$ — below the axis.
Intersections in the third and fourth quadrants.
$\sin^{-1}(0.7) \approx 44.4\degree$ $sin^{- 1} (0.7) \approx 44.4°$ , so:
- Third quadrant: $180\degree + 44.4\degree = 224.4\degree$ .
- Fourth quadrant: $360\degree - 44.4\degree = 315.6\degree$ .
$x \approx 224.4\degree, 315.6\degree$ .

Sketch the trig graph in range.
Draw $y = k$ .
Read off intersections.
Use symmetry to convert principal to all solutions.

How it’s examined

Trig equations appear most years on Paper 4 (3-5 marks), often as part of a longer trig graph question. Examiner reports flag missing the second solution as the recurring lost mark — students give only the calculator's principal value and stop. Always look for ALL solutions in the range.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E8.6); 0580/42 Oct/Nov 2024 — Q19 (trig equation, solve in range); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

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Step-by-step worked examples — Trigonometric Equations

Step-by-step solutions to past-paper-style questions on trigonometric equations, written exactly the way a tutor would explain them at the board.

1Solve $\sin\theta = 0.7$
Extended• Adapted from 0580/42 May/Jun 2024 Q18• sin
Question
Solve $\sin\theta = 0.7$ for $0\degree \le \theta \le 360\degree$ .
Step-by-step solution
1. Step 1
  Principal value.
  $\theta_{1} = \sin^{-1}(0.7) \approx 44.4\degree$
2. Step 2
  Sin is positive in 1st and 2nd quadrants → second solution.
  $\theta_{2} = 180\degree - 44.4\degree = 135.6\degree$
Answer
$\theta \approx 44.4\degree, 135.6\degree$
2Solve $\cos\theta = -0.3$
Extended• cos
Question
Solve $\cos\theta = -0.3$ for $0\degree \le \theta \le 360\degree$ .
Step-by-step solution
1. Step 1
  Principal value.
  $\theta_{1} = \cos^{-1}(-0.3) \approx 107.5\degree$
2. Step 2
  Cos is symmetric about $180\degree$ : second solution = $360\degree - \theta_{1}$ .
  $\theta_{2} = 360\degree - 107.5\degree = 252.5\degree$
Answer
$\theta \approx 107.5\degree, 252.5\degree$
3Solve $\tan\theta = 1$
Extended• tan
Question
Solve $\tan\theta = 1$ for $0\degree \le \theta \le 360\degree$ .
Step-by-step solution
1. Step 1
  Principal value.
  $\theta_{1} = \tan^{-1}(1) = 45\degree$
2. Step 2
  Tan has period $180\degree$ .
  $\theta_{2} = 45\degree + 180\degree = 225\degree$
Answer
$\theta = 45\degree, 225\degree$
4Rearrange before solving
Extended• rearrange
Question
Solve $2\sin\theta - 1 = 0$ for $0\degree \le \theta \le 360\degree$ .
Step-by-step solution
1. Step 1
  Isolate $\sin\theta$ .
  $\sin\theta = \dfrac{1}{2}$
2. Step 2
  Two solutions in the range.
  $\theta = 30\degree, 150\degree$
Answer
$\theta = 30\degree, 150\degree$
5Solve $\sin x = 0.5$ in $[0\degree, 360\degree]$
Core• sin, basic
Question
Solve $\sin x = 0.5$ for $0\degree \le x \le 360\degree$ .
Step-by-step solution
1. Step 1
  Principal value.
  $x_{1} = \sin^{-1}(0.5) = 30\degree$
2. Step 2
  Second solution from sine symmetry: $x_{2} = 180\degree - x_{1}$ .
  $x_{2} = 180\degree - 30\degree = 150\degree$
Answer
$x = 30\degree,\ 150\degree$
Examiner tip
Both solutions are exact, so give exact integer values. Approximating as $30.0\degree$ unnecessarily is harmless but state exact angles where possible.
6Solve $\cos x = -\dfrac{1}{\sqrt{2}}$
Core• Adapted from 0580/22 Feb/Mar 2023 Q18• cos, exact
Question
Solve $\cos x = -\dfrac{1}{\sqrt{2}}$ for $0\degree \le x \le 360\degree$ .
Step-by-step solution
1. Step 1
  $\cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right) = 45\degree$ (acute reference angle).
2. Step 2
  Cosine is negative in the 2nd and 3rd quadrants.
  $x_{1} = 180\degree - 45\degree = 135\degree$
3. Step 3
  Third-quadrant solution.
  $x_{2} = 180\degree + 45\degree = 225\degree$
Answer
$x = 135\degree,\ 225\degree$
Examiner tip
$\dfrac{1}{\sqrt{2}}$ is one of the exact trig values 0580 expects. Memorising the reference angles for $0.5, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}$ saves calculator time and prevents rounding loss.
7Solve $\tan x = \sqrt{3}$
Extended• tan, exact
Question
Solve $\tan x = \sqrt{3}$ for $0\degree \le x \le 360\degree$ .
Step-by-step solution
1. Step 1
  Principal value (exact).
  $x_{1} = \tan^{-1}(\sqrt{3}) = 60\degree$
2. Step 2
  Tan has period $180\degree$ .
  $x_{2} = 60\degree + 180\degree = 240\degree$
Answer
$x = 60\degree,\ 240\degree$
8Solve $\sin(2\theta) = 0.5$
Extended• Adapted from 0580/42 Oct/Nov 2024 Q18• compound argument
Question
Solve $\sin(2\theta) = 0.5$ for $0\degree \le \theta \le 360\degree$ .
Step-by-step solution
1. Step 1
  Let $u = 2\theta$ . The interval becomes $0\degree \le u \le 720\degree$ .
2. Step 2
  Solve $\sin u = 0.5$ . Principal $u = 30\degree$ . Other solutions in $[0\degree, 720\degree]$ : $150\degree, 30\degree + 360\degree = 390\degree, 150\degree + 360\degree = 510\degree$ .
  $u = 30\degree, 150\degree, 390\degree, 510\degree$
3. Step 3
  Divide each by $2$ to recover $\theta$ .
  $\theta = 15\degree, 75\degree, 195\degree, 255\degree$
Answer
$\theta = 15\degree,\ 75\degree,\ 195\degree,\ 255\degree$
Examiner tip
Doubling the variable doubles the interval — find solutions in $[0\degree, 720\degree]$ first, then halve. The 2024 examiner report flags candidates who solve only in $[0\degree, 360\degree]$ and lose two of the four solutions.
9A* quadratic in $\sin\theta$
Challenge• Adapted from 0580/42 May/Jun 2023 Q24• quadratic, challenge
Question
Solve $2\sin^{2}\theta - \sin\theta - 1 = 0$ for $0\degree \le \theta \le 360\degree$ .
Step-by-step solution
1. Step 1
  Let $s = \sin\theta$ . The equation becomes $2s^{2} - s - 1 = 0$ .
2. Step 2
  Factorise.
  $(2s + 1)(s - 1) = 0 \Rightarrow s = -\dfrac{1}{2}\ \text{or}\ s = 1$
3. Step 3
  Case $\sin\theta = 1$ : $\theta = 90\degree$ (only solution in range).
4. Step 4
  Case $\sin\theta = -\dfrac{1}{2}$ : $\theta = 180\degree + 30\degree = 210\degree$ or $360\degree - 30\degree = 330\degree$ .
Answer
$\theta = 90\degree,\ 210\degree,\ 330\degree$
Examiner tip
Substitute first, factorise second, then solve each case. The 2023 mark scheme splits the marks: factorisation, each root for $\sin\theta$ , and each angle in range.
10A* equation with both $\sin$ and $\cos$
Challenge• Adapted from 0580/42 Feb/Mar 2024 Q25• mixed trig, challenge
Question
Solve $\sin\theta = \cos\theta$ for $0\degree \le \theta \le 360\degree$ .
Step-by-step solution
1. Step 1
  Divide both sides by $\cos\theta$ (assuming $\cos\theta \neq 0$ , i.e. $\theta \neq 90\degree, 270\degree$ ).
  $\tan\theta = 1$
2. Step 2
  Principal value.
  $\theta_{1} = \tan^{-1}(1) = 45\degree$
3. Step 3
  Add the period of $\tan$ to find the second solution in range.
  $\theta_{2} = 45\degree + 180\degree = 225\degree$
4. Step 4
  Check the excluded angles: at $\theta = 90\degree$ , $\sin = 1$ but $\cos = 0$ , so the original equation is not satisfied. Same for $\theta = 270\degree$ . No extra solutions.
Answer
$\theta = 45\degree,\ 225\degree$
Examiner tip
Dividing by $\cos\theta$ requires you to check that $\cos\theta \neq 0$ — the 2024 mark scheme awards one mark for stating this check explicitly.

Key Formulae — Trigonometric Equations

The formulae you need to memorise for trigonometric equations on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Second solution for $\sin$
$\text{If } \sin\theta_{1} = k,\text{ then } \theta_{2} = 180\degree - \theta_{1}$
When to use
Range $[0\degree, 360\degree]$ for $\sin\theta = k$ .
Second solution for $\cos$
$\text{If } \cos\theta_{1} = k,\text{ then } \theta_{2} = 360\degree - \theta_{1}$
When to use
Range $[0\degree, 360\degree]$ for $\cos\theta = k$ .
Second solution for $\tan$
$\text{If } \tan\theta_{1} = k,\text{ then } \theta_{2} = \theta_{1} + 180\degree$
When to use
Range $[0\degree, 360\degree]$ for $\tan\theta = k$ .

Key Definitions and Keywords — Trigonometric Equations

Definitions to memorise and the exact keywords mark schemes credit for trigonometric equations answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Solution range
Examiner keyword
The interval of $\theta$ within which solutions are required (e.g. $0\degree \le \theta \le 360\degree$ ).
Quadrant
Examiner keyword
One of the four regions of the trig graph (each $90\degree$ wide). The CAST diagram shows where each ratio is positive.

Common Mistakes and Misconceptions — Trigonometric Equations

The traps other students keep falling into on trigonometric equations questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Giving only the principal value
0580/42 — every series
Why it happens
Calculator returns one angle; student stops.
How to avoid it
After the principal, USE THE SYMMETRY rule to get the second value in range.
✕Including solutions outside the given range
Why it happens
Forgetting to check the bounds.
How to avoid it
Reject any solution below $0\degree$ or above $360\degree$ (or whatever range was specified).
✕Calculator in radian mode
Why it happens
Default mode varies.
How to avoid it
Switch to DEG. Sanity check: $\sin^{-1}(0.5) = 30\degree$ , not $0.524$ .
✕Solving $\sin 2\theta = k$ as if it were $\sin\theta = k$
Why it happens
Treating the multiplied argument as the variable.
How to avoid it
Substitute $u = 2\theta$ , find $u$ in the doubled range, then divide by 2 at the end.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Video lesson

Short walkthrough of the concepts students most often get stuck on.

Trigonometric Equations — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Trigonometric Equations

Short Study Notes in the form of Slides

Short Study Notes — Trigonometric Equations

Detailed Notes

What you’ll learn

How it’s examined

1Solve sin⁡θ=0.7\sin\theta = 0.7sinθ=0.7

2Solve cos⁡θ=−0.3\cos\theta = -0.3cosθ=−0.3

3Solve tan⁡θ=1\tan\theta = 1tanθ=1

4Rearrange before solving

5Solve sin⁡x=0.5\sin x = 0.5sinx=0.5 in [0°,360°][0\degree, 360\degree][0°,360°]

6Solve cos⁡x=−12\cos x = -\dfrac{1}{\sqrt{2}}cosx=−2​1​

7Solve tan⁡x=3\tan x = \sqrt{3}tanx=3​

8Solve sin⁡(2θ)=0.5\sin(2\theta) = 0.5sin(2θ)=0.5

9A* quadratic in sin⁡θ\sin\thetasinθ

10A* equation with both sin⁡\sinsin and cos⁡\coscos

Second solution for sin⁡\sinsin

Second solution for cos⁡\coscos

Second solution for tan⁡\tantan

Solution range

Quadrant

✕Giving only the principal value

✕Including solutions outside the given range

✕Calculator in radian mode

✕Solving sin⁡2θ=k\sin 2\theta = ksin2θ=k as if it were sin⁡θ=k\sin\theta = ksinθ=k

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Trigonometric Equations — frequently asked questions

1Solve $\sin\theta = 0.7$

2Solve $\cos\theta = -0.3$

3Solve $\tan\theta = 1$

5Solve $\sin x = 0.5$ in $[0\degree, 360\degree]$

6Solve $\cos x = -\dfrac{1}{\sqrt{2}}$

7Solve $\tan x = \sqrt{3}$

8Solve $\sin(2\theta) = 0.5$

9A* quadratic in $\sin\theta$

10A* equation with both $\sin$ and $\cos$

Second solution for $\sin$

Second solution for $\cos$

Second solution for $\tan$

✕Solving $\sin 2\theta = k$ as if it were $\sin\theta = k$