Solving Trigonometric Equations

This lesson covers techniques for solving trigonometric equations at A-Level, including equations involving multiple angles, equations reducible to quadratics, equations requiring the use of identities, and finding general solutions. This is one of the most frequently examined topics in A-Level trigonometry.

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General Approach

1. Reduce to a single trigonometric function using identities if needed.
2. Find the principal value using inverse trig functions or exact values.
3. Find all solutions in the given range using the CAST diagram or symmetry.
4. Check your answers by substituting back.

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Basic Equations

Equations of the form sin x = k, cos x = k, tan x = k

Example 1: Solve sin x = 1/2 for 0° ≤ x ≤ 360°.

Principal value: x = arcsin(1/2) = 30°.

sin is positive in quadrants 1 and 2 (from CAST diagram):

x = 30° or x = 180° − 30° = 150°

Solutions: x = 30°, 150°

Example 2: Solve cos x = −√3/2 for 0 ≤ x ≤ 2π.

Principal value: arccos(√3/2) = π/6. Since cos is negative, we are in quadrants 2 and 3:

x = π − π/6 = 5π/6 or x = π + π/6 = 7π/6

Solutions: x = 5π/6, 7π/6

Example 3: Solve tan x = −1 for 0° ≤ x ≤ 360°.

arctan(1) = 45°. tan is negative in quadrants 2 and 4:

x = 180° − 45° = 135° or x = 360° − 45° = 315°

Solutions: x = 135°, 315°

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Equations with Multiple Angles

For equations involving sin(kx), cos(kx), or tan(kx), adjust the range for the substitution u = kx.

Example 4: Solve sin 2x = √3/2 for 0° ≤ x ≤ 360°.

Let u = 2x. Since 0° ≤ x ≤ 360°, we have 0° ≤ u ≤ 720°.

sin u = √3/2, so the reference angle is 60°.

In 0° ≤ u ≤ 720°:

u = 60°, 120°, 420°, 480°

Therefore:

x = 30°, 60°, 210°, 240°

Example 5: Solve cos 3x = 1/2 for 0 ≤ x ≤ 2π.

Let u = 3x. Since 0 ≤ x ≤ 2π, we have 0 ≤ u ≤ 6π.

cos u = 1/2, so the reference angle is π/3.

In 0 ≤ u ≤ 6π:

u = π/3, 5π/3, π/3 + 2π, 5π/3 + 2π, π/3 + 4π, 5π/3 + 4π
u = π/3, 5π/3, 7π/3, 11π/3, 13π/3, 17π/3

Therefore:

x = π/9, 5π/9, 7π/9, 11π/9, 13π/9, 17π/9

Example 6: Solve tan(x + 45°) = √3 for 0° ≤ x ≤ 360°.

Let u = x + 45°. Since 0° ≤ x ≤ 360°, we have 45° ≤ u ≤ 405°.

tan u = √3, reference angle is 60°.

tan is positive in quadrants 1 and 3:

u = 60°, 240°, or 60° + 360° = 420° (still need to check if in range)

u = 60°: x = 15° ✓ u = 240°: x = 195° ✓ u = 420°: x = 375° > 360° ✗

Solutions: x = 15°, 195°

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Equations Reducible to Quadratics

Many trig equations can be transformed into quadratic equations by making a substitution.

Example 7: Solve 2sin²x − sin x − 1 = 0 for 0° ≤ x ≤ 360°.

Let u = sin x:

2u² − u − 1 = 0
(2u + 1)(u − 1) = 0
u = −1/2 or u = 1

sin x = −1/2: x = 210°, 330° sin x = 1: x = 90°

Solutions: x = 90°, 210°, 330°

Example 8: Solve 2cos²x + cos x − 1 = 0 for 0 ≤ x ≤ 2π.

Let u = cos x:

2u² + u − 1 = 0
(2u − 1)(u + 1) = 0
u = 1/2 or u = −1

cos x = 1/2: x = π/3, 5π/3 cos x = −1: x = π

Solutions: x = π/3, π, 5π/3

Example 9: Solve 4sin²x − 1 = 0 for 0° ≤ x ≤ 360°.

sin²x = 1/4
sin x = ±1/2

sin x = 1/2: x = 30°, 150° sin x = −1/2: x = 210°, 330°

Solutions: x = 30°, 150°, 210°, 330°

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Using Identities to Solve Equations

Using sin²x + cos²x = 1

Example 10: Solve 3sin²x + sin x cos x = 0 for 0° ≤ x ≤ 360°.

Factorise:

sin x(3sin x + cos x) = 0

sin x = 0: x = 0°, 180°, 360°

3sin x + cos x = 0:

3sin x = −cos x
tan x = −1/3
x = 180° + arctan(−1/3) = 180° − 18.43° = 161.57° or 360° − 18.43° = 341.57°

Solutions: x = 0°, 161.6°, 180°, 341.6°, 360° (to 1 d.p.)

Using 1 + tan²x = sec²x

Example 11: Solve 2tan²x + sec x = 1 for 0 ≤ x ≤ 2π.

Replace tan²x with sec²x − 1:

2(sec²x − 1) + sec x = 1
2sec²x − 2 + sec x = 1
2sec²x + sec x − 3 = 0

Let u = sec x:

2u² + u − 3 = 0
(2u + 3)(u − 1) = 0
u = −3/2 or u = 1

sec x = 1: cos x = 1, so x = 0 or x = 2π sec x = −3/2: cos x = −2/3, so x = arccos(−2/3) = 2.301 or x = 2π − 2.301 = 3.982

Solutions: x = 0, 2.30, 3.98, 2π (to 3 s.f.)

Using sin 2x = 2sin x cos x

Example 12: Solve sin 2x + cos x = 0 for 0 ≤ x ≤ 2π.

2sin x cos x + cos x = 0
cos x(2sin x + 1) = 0

cos x = 0: x = π/2, 3π/2 sin x = −1/2: x = 7π/6, 11π/6

Solutions: x = π/2, 7π/6, 3π/2, 11π/6

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General Solutions

Sometimes you are asked for the general solution — all solutions, not just those in a finite range.

Equation	General Solution
sin x = k	x = nπ + (−1)ⁿ arcsin k, n ∈ ℤ
cos x = k	x = 2nπ ± arccos k, n ∈ ℤ
tan x = k	x = nπ + arctan k, n ∈ ℤ

Example 13: Find the general solution of sin x = √3/2.

arcsin(√3/2) = π/3
General solution: x = nπ + (−1)ⁿ(π/3), where n is any integer

This gives: ..., −2π/3, π/3, 2π/3, 7π/3, ...

Alternatively: x = π/3 + 2nπ or x = 2π/3 + 2nπ (this alternative form is often clearer).

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Summary

• For basic equations, find the principal value and use the CAST diagram for other solutions.
• For multiple angle equations (sin kx = ...), extend the range and divide solutions by k.
• For quadratic-type equations, use a substitution (let u = sin x, etc.) and solve the quadratic.
• Use identities (Pythagorean, double angle) to reduce equations to a single trig function.
• Never divide by a trig function — always factorise to avoid losing solutions.
• For general solutions, use the standard formulae with n ∈ ℤ.

Exam Tip: The most common error in solving trig equations is missing solutions. Always check the number of solutions you would expect (based on the graph), and verify each solution is in the required range. When the equation involves sin 2x, remember to extend the range for u = 2x before finding solutions. Show all your working clearly, including the principal value and how you found additional solutions.

A-Level Deep Dive: Solving Trigonometric Equations

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section E (Trigonometry) covers simple trigonometric equations in a given interval, including quadratic equations in $\sin$sin, $\cos$cos and $\tan$tan and equations involving multiples of the unknown angle (refer to the official specification document for exact wording). Equation-solving is the terminal skill in Section E — it draws on every preceding sub-strand: exact values, the unit-circle definitions, the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$sin2θ+cos2θ=1, double-angle formulae, the harmonic form $R\sin(\theta + \alpha)$Rsin(θ+α), and the periodicities $\sin(\theta + 2\pi) = \sin\theta$sin(θ+2π)=sinθ, $\tan(\theta + \pi) = \tan\theta$tan(θ+π)=tanθ. The AQA formula booklet provides the compound-angle and double-angle identities; the Pythagorean identity and the periodicities are not listed and must be memorised. Solving trigonometric equations is examined on Paper 2 (Pure) and recurs in Paper 3 (Statistics and Mechanics) wherever simple harmonic motion appears.

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Worked example with full mark scheme

Question (8 marks): Solve the equation

$\sin 2x + \sin x = 0$sin2x+sinx=0

for $x \in [0, 2\pi]$x∈[0,2π], giving all solutions in exact form.

Solution with mark scheme:

Step 1 — apply the double-angle identity.

The identity $\sin 2x = 2 \sin x \cos x$sin2x=2sinxcosx is in the formula booklet. Substitute:

$2 \sin x \cos x + \sin x = 0$2sinxcosx+sinx=0

M1 — correct application of the double-angle formula. A common slip is to write $\sin 2x = 2 \sin x$sin2x=2sinx (missing the $\cos x$cosx factor), which collapses the problem incorrectly and earns nothing.

Step 2 — factorise.

$\sin x \, (2 \cos x + 1) = 0$sinx(2cosx+1)=0

M1 — factorising out $\sin x$sinx. Crucially, do not divide by $\sin x$sinx — that loses the $\sin x = 0$sinx=0 family of solutions. Examiners specifically watch for this; a candidate who writes "divide by $\sin x$sinx to get $2\cos x + 1 = 0$2cosx+1=0" forfeits up to four marks because half the solution set vanishes.

A1 — correct factorised form.

Step 3 — solve each factor on $[0, 2\pi]$[0,2π].

Branch 1: $\sin x = 0$sinx=0 gives $x = 0, \pi, 2\pi$x=0,π,2π.

A1 — all three values within the closed interval. Missing $2\pi$2π (because the candidate uses $[0, 2\pi)$[0,2π) instead of the printed $[0, 2\pi]$[0,2π]) is a typical slip.

Branch 2: $2 \cos x + 1 = 0 \implies \cos x = -\tfrac{1}{2}$2cosx+1=0⟹cosx=−21​.

The principal value is $x = \arccos(-\tfrac{1}{2}) = \tfrac{2\pi}{3}$x=arccos(−21​)=32π​. Cosine is negative in the second and third quadrants, so the second solution on $[0, 2\pi]$[0,2π] is $x = 2\pi - \tfrac{2\pi}{3} = \tfrac{4\pi}{3}$x=2π−32π​=34π​.

M1 — correct principal value.

A1 — both values $\tfrac{2\pi}{3}$32π​ and $\tfrac{4\pi}{3}$34π​.

Step 4 — collate.

$x \in \left\{0, \, \tfrac{2\pi}{3}, \, \pi, \, \tfrac{4\pi}{3}, \, 2\pi\right\}$x∈{0,32π​,π,34π​,2π}

A1 — complete solution set, in exact form, in increasing order.

Total: 8 marks (M3 A5). Most marks here are accuracy marks because the manipulation is short — the examiner is testing whether you find every root in the interval.

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Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): Solve $\sin\!\left(2x + \tfrac{\pi}{4}\right) = \tfrac{1}{2}$sin(2x+4π​)=21​ for $x \in [0, \pi]$x∈[0,π], giving exact solutions.

Mark scheme decomposition by AO:

• M1 (AO1.1a) — substituting $u = 2x + \tfrac{\pi}{4}$u=2x+4π​ and adjusting the interval. When $x \in [0, \pi]$x∈[0,π], $u \in [\tfrac{\pi}{4}, 2\pi + \tfrac{\pi}{4}] = [\tfrac{\pi}{4}, \tfrac{9\pi}{4}]$u∈[4π​,2π+4π​]=[4π​,49π​].
• M1 (AO1.1b) — finding the principal value: \sin u = \tfrac{1}{2}\) gives u = \tfrac{\pi}{6}$as the principal value, but$astheprincipalvalue,but\tfrac{\pi}{6} \notin [\tfrac{\pi}{4}, \tfrac{9\pi}{4}]$, so the smallest valid value comes from the second-quadrant solution$,sothesmallestvalidvaluecomesfromthesecond−quadrantsolutionu = \pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}$.
• A1 (AO1.1b) — full set in $u$u: $u = \tfrac{5\pi}{6}, \tfrac{\pi}{6} + 2\pi = \tfrac{13\pi}{6}, \tfrac{5\pi}{6} + 2\pi = \tfrac{17\pi}{6}$u=65π​,6π​+2π=613π​,65π​+2π=617π​. The last is outside $[\tfrac{\pi}{4}, \tfrac{9\pi}{4}]$[4π​,49π​], so reject. Valid $u$u values: $\tfrac{5\pi}{6}, \tfrac{13\pi}{6}$65π​,613π​.
• M1 (AO2.5) — back-substituting $x = \tfrac{u - \pi/4}{2}$x=2u−π/4​.
• A1 (AO1.1b) — $x = \tfrac{1}{2}(\tfrac{5\pi}{6} - \tfrac{\pi}{4}) = \tfrac{1}{2} \cdot \tfrac{10\pi - 3\pi}{12} = \tfrac{7\pi}{24}$x=21​(65π​−4π​)=21​⋅1210π−3π​=247π​.
• A1 (AO1.1b) — $x = \tfrac{1}{2}(\tfrac{13\pi}{6} - \tfrac{\pi}{4}) = \tfrac{1}{2} \cdot \tfrac{26\pi - 3\pi}{12} = \tfrac{23\pi}{24}$x=21​(613π​−4π​)=21​⋅1226π−3π​=2423π​.

Total: 6 marks split AO1 = 4, AO2 = 2. The AO2 mark is for correctly transforming the interval — the single biggest source of mark loss on transformed-argument equations.

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Synoptic links

Connects to:

1. All of Section E (Trigonometry): equation-solving uses every prior identity. Pythagorean substitution turns $\cos^2 \theta$cos2θ into $1 - \sin^2 \theta$1−sin2θ to expose a hidden quadratic; double-angle formulae linearise products; the harmonic form $R\sin(\theta + \alpha)$Rsin(θ+α) collapses $a \sin \theta + b \cos \theta$asinθ+bcosθ into a single sinusoid amenable to inversion. Solving is the proving ground for fluency in identities.

2. Calculus (Section G) — stationary points: finding $\frac{dy}{dx} = 0$dxdy​=0 for $y = \sin x \cos x$y=sinxcosx requires solving $\cos 2x = 0$cos2x=0. Every locating-stationary-points question on a trigonometric function reduces to a trigonometric equation in a given interval.

3. Harmonic form combinations (Section E continued): equations like $3 \sin \theta + 4 \cos \theta = 5$3sinθ+4cosθ=5 are unsolvable by direct inversion but trivial after the harmonic conversion $R\sin(\theta + \alpha) = 5$Rsin(θ+α)=5. The synoptic skill is recognising when raw inversion will fail and a harmonic re-write is needed.

4. Complex roots (Further Maths) — $e^{i\theta} = 1$eiθ=1: the equation $z^n = 1$zn=1 has roots $z_k = e^{2\pi i k / n}$zk​=e2πik/n, which on the Argand diagram correspond to solving $\cos\!\left(\tfrac{2\pi k}{n}\right) + i \sin\!\left(\tfrac{2\pi k}{n}\right) = 1$cos(n2πk​)+isin(n2πk​)=1. The same periodicity argument that finds all $x \in [0, 2\pi]$x∈[0,2π] from a principal value of $\sin x = c$sinx=c generalises to enumerating roots of unity.

5. Simple harmonic motion (Paper 3 — Mechanics): the SHM displacement $x(t) = A \cos(\omega t + \phi)$x(t)=Acos(ωt+ϕ) requires solving $x(t) = x_0$x(t)=x0​ for the times of given displacement. Adjusting the interval for the transformed argument $\omega t + \phi$ωt+ϕ is the same skill drilled in pure trigonometry.

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Mark-scheme literacy

Equation-solving questions on AQA Paper 2 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Applying identities, finding principal values, listing solutions
AO2 (reasoning / interpretation)	25–35%	Adjusting intervals for transformed arguments, justifying which roots to keep, choosing between Pythagorean and double-angle routes
AO3 (problem-solving)	10–20%	Selecting harmonic-form conversion when direct inversion fails; recognising hidden quadratics; equations from a modelling context

Examiner-rewarded phrasing: "the principal value is $x = \arccos(c)$x=arccos(c), and since $\cos$cos is positive in the first and fourth quadrants, the second solution is $2\pi - \arccos(c)$2π−arccos(c)"; "when $x \in [0, 2\pi]$x∈[0,2π], the substituted variable $u = 2x$u=2x runs through $[0, 4\pi]$[0,4π], so we take principal value plus periodic copies up to $4\pi$4π"; "rejecting $u = \tfrac{17\pi}{6}$u=617π​ as it lies outside the transformed interval". Phrases that lose marks: writing only the principal value as the "answer"; using degrees when the question prints radians (or vice versa); listing solutions outside the printed interval without rejecting them.