Can I use the R-formula for questions?

No, the R-formula is A-Level content and out of scope for 0606. Only equations are tested, which you solve by rearranging to a equation.

Solving Trigonometric Equations

Q: Do I need to give exact solutions or decimal approximations?

Follow question instructions. Use the same unit as the given domain. For Paper 1 (non-calculator), solutions will be exact standard angles, while Paper 2 allows decimal approximations to 3 significant figures unless stated otherwise.

Q: How do I avoid missing solutions in the given interval?

First find the principal solution, then use periodicity and quadrant symmetry rules for each trigonometric function to find all other solutions that fit the domain, before checking for extraneous solutions.

CIE IGCSE Additional Mathematics· 10.5· 25 min read

1. Linear Trigonometric Equations Over a Specified Domain★★☆☆☆⏱ 7 min

Linear trigonometric equations have the form $k \times trig (θ) = c$ , where $k$ and $c$ are constants, and $trig (θ)$ 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.

📘 Definition

Principal Solution

The first (least non-negative) solution of a trigonometric equation, lying in $[0^{\circ}, 36 0^{\circ})$ for degrees or $[0, 2 π)$ for radians. This is not the same as a calculator's 'principal value' (the inverse-function output range).

📐 Worked Example

Solve $2 sin θ + 1 = 0$ for $θ$ in the interval $0^{\circ} \leq θ < 36 0^{\circ}$ .

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: $θ = 21 0^{\circ}, 33 0^{\circ}$

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.

2. Solving Equations Using Reciprocal Identities★★★☆☆⏱ 7 min

✓ Calculator OK

Reciprocal identities (e.g. $sec θ = \frac{1}{c o s θ}$ , $cot θ = \frac{1}{t a n θ}$ ) 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.

📐 Worked Example

Solve $4 cot θ = tan θ$ for $0 \leq θ < 2 π$ .

1
Rewrite cot using reciprocal identity, noting $tan θ \neq = 0$ :
$4 \times \frac{1}{\tan\theta} = \tan\theta$
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: $1.107$ , $1.107 + π \approx 4.249$ , $π - 1.107 \approx 2.034$ , $2 π - 1.107 \approx 5.176$
5
Verify none of these make $tan θ = 0$ , so all are valid. Final solutions (3 s.f.): $1.11, 2.03, 4.25, 5.18$ radians

3. Solving Quadratic Trigonometric Equations Using Pythagorean Identities★★★★☆⏱ 8 min

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

📘 Definition

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 $2 sec^{2} θ + tan θ - 3 = 0$ for $0^{\circ} \leq θ < 36 0^{\circ}$ .

1
Replace $sec^{2} θ$ 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 $2 tan θ - 1 = 0$ : $tan θ = 0.5$ . Solutions: $26. 6^{\circ}$ , $206. 6^{\circ}$
5
Solve second factor $tan θ + 1 = 0$ : $tan θ = - 1$ . Solutions: $13 5^{\circ}$ , $31 5^{\circ}$
6
Verify all solutions are valid (sec is defined for all values here). Final solutions: $26. 6^{\circ}, 13 5^{\circ}, 206. 6^{\circ}, 31 5^{\circ}$

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 $sin θ$ greater than 1 or less than -1).

4. Solving $a\sin\theta + b\cos\theta = 0$ Equations★★★☆☆⏱ 5 min

For equations of the form $a sin θ + b cos θ = 0$ , you can rearrange and divide by $cos θ$ (as long as $cos θ \neq = 0$ ) to get a tan equation, which you can solve using standard methods. The R-formula for $a sin θ + b cos θ = c$ (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 $3 sin θ - 2 cos θ = 0$ for $0^{\circ} \leq θ < 36 0^{\circ}$ .

1
Rearrange the equation to group terms:
$3\sin\theta = 2\cos\theta$
2
Divide both sides by $cos θ$ , noting $cos θ \neq = 0$ (no solutions lost as $cos θ = 0$ would make LHS=±3, RHS=0):
$3\tan\theta = 2 \implies \tan\theta = \frac{2}{3} \approx 0.6667$
3
Find solutions: $tan^{- 1} (2/3) \approx 33. 7^{\circ}$ , $33. 7^{\circ} + 18 0^{\circ} = 213. 7^{\circ}$
4
Verify both solutions are valid. Final solutions: $33. 7^{\circ}, 213. 7^{\circ}$

5. Solving Equations with Multiple or Shifted Angles★★★☆☆⏱ 8 min

Many exam equations act on a multiple or shifted angle, such as $sin 2 x$ , $cos 3 θ$ or $tan (θ - 3 0^{\circ})$ . To solve $trig (b x + d) = k$ , substitute $u = b x + d$ , 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 $x$ .

📐 Worked Example

Solve $sin 2 x = \frac{3}{2}$ for $0^{\circ} \leq x < 36 0^{\circ}$ .

1
Substitute $u = 2 x$ and transform the domain. Multiplying $0^{\circ} \leq x < 36 0^{\circ}$ through by 2 gives:
$0^\circ \leq u < 720^\circ$
2
Solve $sin u = \frac{3}{2}$ . Reference angle $α = 6 0^{\circ}$ ; sine is positive in the 1st and 2nd quadrants, giving the first-revolution solutions:
$u = 60^\circ,\ 120^\circ$
3
Add $36 0^{\circ}$ to reach the second revolution, keeping only values below $72 0^{\circ}$ :
$u = 420^\circ,\ 480^\circ$
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 $sin (2 θ + \frac{π}{6}) = \frac{1}{2}$ for $0 \leq θ < 2 π$ . Give exact answers in terms of $π$ .

1
Substitute $u = 2 θ + \frac{π}{6}$ and transform the domain. As $θ$ runs over $0 \leq θ < 2 π$ , $u$ runs from $\frac{π}{6}$ up to (but not including) $2 (2 π) + \frac{π}{6} = \frac{25 π}{6}$ :
$\frac{\pi}{6} \leq u < \frac{25\pi}{6}$
2
Solve $sin u = \frac{1}{2}$ . Reference angle $α = \frac{π}{6}$ ; sine is positive in the 1st and 2nd quadrants, giving base solutions (both lie in the domain, which starts at $\frac{π}{6}$ ):
$u = \frac{\pi}{6},\ \frac{5\pi}{6}$
3
Add $2 π$ for the next period; the value $\frac{25 π}{6}$ is the excluded upper limit, so stop here:
$u = \frac{13\pi}{6},\ \frac{17\pi}{6}$
4
Reverse the substitution with $θ = \frac{u - \frac{π}{6}}{2}$ 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.

6. Common Pitfalls

Wrong move:

Only giving the principal solution and forgetting other solutions in the domain

Why:

Trigonometric functions are periodic, so there are often multiple solutions in a given interval

Correct move:

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 move:

Dividing both sides of an equation by a trigonometric function without checking for solutions where that function is zero

Why:

This removes valid solutions where the function equals zero

Correct move:

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 move:

Using the R-formula to solve $a sin θ + b cos θ = c$ for $c \neq = 0$

Why:

The R-formula is A-Level content, out of scope for 0606, and these equations will not appear on the exam

Correct move:

Only solve $a sin θ + b cos θ = 0$ by rearranging to get a tan equation as specified in the syllabus

Wrong move:

Forgetting to check for extraneous solutions after squaring both sides or using identities

Why:

Squaring both sides can introduce solutions that do not satisfy the original equation

Correct move:

Substitute every solution you find back into the original equation to confirm it is valid before including it in your final answer

Wrong move:

Mixing up degrees and radians when solving equations

Why:

The question specifies the unit in the domain, and using the wrong unit leads to incorrect answers, especially on calculator papers

Correct move:

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

7. Quick Reference Cheatsheet

Trigonometric Function	Symmetry Rule for Solutions (0° ≤ θ < 360°)	Period
$sin θ = k$	If k positive: $α, 18 0^{\circ} - α$ ; If k negative: $18 0^{\circ} + α, 36 0^{\circ} - α$ where $α = sin^{- 1} (∣ k ∣)$	360° / 2π
$cos θ = k$	If k positive: $α, 36 0^{\circ} - α$ ; If k negative: $18 0^{\circ} - α, 18 0^{\circ} + α$ where $α = cos^{- 1} (∣ k ∣)$	360° / 2π
$tan θ = k$	If k positive: $α, 18 0^{\circ} + α$ ; If k negative: $18 0^{\circ} - α, 36 0^{\circ} - α$ where $α = tan^{- 1} (∣ k ∣)$	180° / π
Reciprocal Identities	$cosec θ = 1/ sin θ$ , $sec θ = 1/ cos θ$ , $cot θ = 1/ tan θ$
Pythagorean Identities	$sin^{2} θ + cos^{2} θ = 1$ , $1 + tan^{2} θ = sec^{2} θ$ , $1 + cot^{2} θ = cosec^{2} θ$

8. Frequently Asked

Do I need to give exact solutions or decimal approximations?

Follow question instructions. Use the same unit as the given domain. For Paper 1 (non-calculator), solutions will be exact standard angles, while Paper 2 allows decimal approximations to 3 significant figures unless stated otherwise.

How do I avoid missing solutions in the given interval?

First find the principal solution, then use periodicity and quadrant symmetry rules for each trigonometric function to find all other solutions that fit the domain, before checking for extraneous solutions.

Can I use the R-formula for $a sin θ + b cos θ = c$ questions?

No, the R-formula is A-Level content and out of scope for 0606. Only $a sin θ + b cos θ = 0$ equations are tested, which you solve by rearranging to a $tan θ$ equation.

Going deeper

study_guideTrigonometric Identities (0606 U10)
practice_setTrigonometric Equations Past Paper Questions

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.