Solving Trigonometric Equations

This lesson brings together all the techniques for solving trigonometric equations at A-Level (9MA0): finding solutions in given ranges, handling multiple solutions, using identities to simplify, and understanding general solutions.

---

Strategy for Solving Trig Equations

1. Simplify using identities if needed (to get a single trig function).
2. Rearrange to isolate the trig function.
3. Find the principal value using inverse trig or exact values.
4. Find all solutions in the given range using CAST diagram or graph.

---

The CAST Diagram

The CAST diagram tells you in which quadrants each trig function is positive:

Quadrant	Angle range	Positive functions
1st	0 to pi/2	All (sin, cos, tan)
2nd	pi/2 to pi	Sin only
3rd	pi to 3pi/2	Tan only
4th	3pi/2 to 2pi	Cos only

---

Solving Basic Equations

Worked Example 1

Solve sin(x) = sqrt(3)/2 for 0 <= x <= 2pi.

Principal value: x = pi/3

Using CAST: sin is positive in Q1 and Q2.

Q1: x = pi/3 Q2: x = pi - pi/3 = 2pi/3

Solutions: x = pi/3, 2pi/3

Worked Example 2

Solve cos(x) = -1/sqrt(2) for 0 <= x <= 2pi.

Principal value: cos^(-1)(1/sqrt(2)) = pi/4

cos is negative in Q2 and Q3.

Q2: x = pi - pi/4 = 3pi/4 Q3: x = pi + pi/4 = 5pi/4

Solutions: x = 3pi/4, 5pi/4

Worked Example 3

Solve tan(x) = -1 for 0 <= x < 2pi.

Principal value: arctan(1) = pi/4

tan is negative in Q2 and Q4.

Q2: x = pi - pi/4 = 3pi/4 Q4: x = 2pi - pi/4 = 7pi/4

Solutions: x = 3pi/4, 7pi/4

---

Solving Equations with Transformed Arguments

Worked Example 4

Solve sin(2x) = 0.5 for 0 <= x <= pi.

Let u = 2x. When x is in [0, pi], u is in [0, 2pi].

sin(u) = 0.5

u = pi/6 or u = 5pi/6

2x = pi/6 or 2x = 5pi/6

x = pi/12 or x = 5pi/12

Solutions: x = pi/12, 5pi/12

Worked Example 5

Solve cos(3x - pi/4) = 0 for 0 <= x <= pi.

Let u = 3x - pi/4. When x is in [0, pi], u is in [-pi/4, 3pi - pi/4] = [-pi/4, 11pi/4].

cos(u) = 0

u = pi/2, 3pi/2, 5pi/2

(Only those values of u in [-pi/4, 11pi/4])

u = pi/2: 3x - pi/4 = pi/2, 3x = 3pi/4, x = pi/4 u = 3pi/2: 3x - pi/4 = 3pi/2, 3x = 7pi/4, x = 7pi/12 u = 5pi/2: 3x - pi/4 = 5pi/2, 3x = 11pi/4, x = 11pi/12

Solutions: x = pi/4, 7pi/12, 11pi/12

---

Using Identities to Reduce Equations

Worked Example 6

Solve 6sin²(x) + cos(x) - 5 = 0 for 0 <= x <= 2pi.

Replace sin²(x) = 1 - cos²(x):

6(1 - cos²(x)) + cos(x) - 5 = 0

6 - 6cos²(x) + cos(x) - 5 = 0

-6cos²(x) + cos(x) + 1 = 0

6cos²(x) - cos(x) - 1 = 0

(3cos(x) + 1)(2cos(x) - 1) = 0

cos(x) = -1/3 or cos(x) = 1/2

cos(x) = 1/2: x = pi/3, 5pi/3

cos(x) = -1/3: x = arccos(-1/3) ≈ 1.911, and x = 2pi - 1.911 ≈ 4.373

Solutions: x = pi/3, 1.911, 4.373, 5pi/3 (to 3 d.p. where needed)

Worked Example 7

Solve 2tan²(x) - 3sec(x) = 0 for 0 <= x < 2pi.

Using tan²(x) = sec²(x) - 1:

2(sec²(x) - 1) - 3sec(x) = 0

2sec²(x) - 2 - 3sec(x) = 0

2sec²(x) - 3sec(x) - 2 = 0

(2sec(x) + 1)(sec(x) - 2) = 0

sec(x) = -1/2 or sec(x) = 2

sec(x) = -1/2: cos(x) = -2, impossible (|cos| <= 1)

sec(x) = 2: cos(x) = 1/2

x = pi/3, 5pi/3

Solutions: x = pi/3, 5pi/3

---

General Solutions

Sometimes you need to give the general solution (for all x, not just in a specific range).

Equation	General solution
sin(x) = k	x = arcsin(k) + 2npi or x = pi - arcsin(k) + 2npi
cos(x) = k	x = ± arccos(k) + 2npi
tan(x) = k	x = arctan(k) + npi

where n is any integer.

Worked Example 8

Find the general solution of tan(x) = sqrt(3).

arctan(sqrt(3)) = pi/3

General solution: x = pi/3 + npi where n is any integer.

Worked Example 9

Find the general solution of 2sin(x) + 1 = 0.

sin(x) = -1/2

arcsin(-1/2) = -pi/6

General solution: x = -pi/6 + 2npi or x = pi + pi/6 + 2npi = 7pi/6 + 2npi

More compactly: x = (-1)^n x (-pi/6) + npi or equivalently x = -pi/6 + 2npi or x = 7pi/6 + 2npi.

---

Equations Requiring Multiple Techniques

Worked Example 10

Solve sin(x) + sin(2x) = 0 for 0 <= x < 2pi.

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

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

sin(x) = 0: x = 0, pi

cos(x) = -1/2: x = 2pi/3, 4pi/3

Solutions: x = 0, 2pi/3, pi, 4pi/3

---

Exam Tips

Tip	Detail
Range of solutions	Always note the required range carefully; if it says 0 <= x <= 2pi you need endpoints too
Do not lose solutions	Factor rather than divide; check all quadrants
Transformed arguments	Adjust the range for the substituted variable before finding solutions
State exact values	Use pi/6, pi/3 etc. when possible rather than decimals
Show full working	Write the CAST analysis or state which quadrants give positive/negative values
General solutions	Use n for integer, and state "where n is any integer"

---

Practice Problems

1. Solve 2cos(x) + 1 = 0 for 0 <= x <= 2pi. [Answer: 2pi/3, 4pi/3]
2. Solve sin(2x) = sin(x) for 0 <= x < 2pi. [Answer: 0, pi/3, pi, 5pi/3]
3. Solve 2cos²(x) + 3sin(x) = 3 for 0 <= x <= 2pi. [Answer: pi/6, pi/2, 5pi/6]
4. Find the general solution of cos(x) = sqrt(3)/2. [Answer: x = ± pi/6 + 2npi]
5. Solve sec²(x) = 4tan(x) - 2 for 0 <= x < 2pi. [Answer: x = arctan(1) = pi/4, 5pi/4 and x = arctan(3) ≈ 1.249, 4.391]

---

Summary

• Use CAST diagram or graphs to find all solutions in a given range.
• For equations with multiple trig functions, use identities to reduce to one function.
• For transformed arguments like sin(2x + pi/4), adjust the range accordingly.
• General solutions use n x pi or 2npi patterns.
• Always factor rather than divide, check for extraneous solutions, and verify your answers.

A-Level Deep Dive: Solving Trigonometric Equations

Spec mapping

Edexcel 9MA0 specification section 5 — Trigonometry, sub-strand on solving trigonometric equations 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). Solving equations is the terminal skill of section 5: it is the point at which every other strand — exact values, the unit circle and CAST diagram, the Pythagorean identities $\sin^2\theta + \cos^2\theta = 1$sin2θ+cos2θ=1, $1 + \tan^2\theta = \sec^2\theta$1+tan2θ=sec2θ, the compound-angle formulae, the double-angle formulae, the small-angle approximations, and the harmonic form $R\sin(\theta + \alpha)$Rsin(θ+α) — gets operationalised. A typical 9MA0 Paper 2 question chains two or three of those tools before the final "solve for $\theta$θ" step.

Synoptic reach beyond section 5. Calculus questions on Paper 2 frequently end with "find the values of $x$x for which $\frac{dy}{dx} = 0$dxdy​=0" where $y$y is a trigonometric expression — that final step is solving a trig equation. In Paper 3 Mechanics, simple harmonic motion solutions of the form $x = A\cos(\omega t + \phi)$x=Acos(ωt+ϕ) require solving $\cos(\omega t + \phi) = k$cos(ωt+ϕ)=k for $t$t to find when the particle reaches a given displacement. The skill of "find all solutions in the given interval" is therefore a recurring cross-paper demand, not a section-5 niche.

The Edexcel formula booklet provides the compound-angle, double-angle and harmonic-form identities, but it does not tell you which identity to substitute or how to organise the principal-plus-symmetry argument. Those decisions are the marks.

---

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 — replace the double-angle term using the identity $\sin 2x = 2\sin x \cos x$sin2x=2sinxcosx.

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

M1 — correct use of the double-angle identity for $\sin 2x$sin2x. A frequent error is to write $\sin 2x = 2\sin x$sin2x=2sinx, treating the $2$2 as a coefficient — this loses both M1 and the entire downstream chain.

Step 2 — factorise rather than divide.

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

M1 — extracting the common factor $\sin x$sinx. The single most-marked error on equations of this shape is dividing both sides by $\sin x$sinx. Doing so loses every solution where $\sin x = 0$sinx=0. Examiners flag this as a serious method error.

Step 3 — set each factor equal to zero.

Either $\sin x = 0$sinx=0 or $2\cos x + 1 = 0$2cosx+1=0, i.e. $\cos x = -\tfrac{1}{2}$cosx=−21​.

A1 — both factor equations stated.

Step 4 — solve $\sin x = 0$sinx=0 on $[0, 2\pi]$[0,2π].

The solutions are $x = 0, \pi, 2\pi$x=0,π,2π.

A1 — all three solutions, with the endpoints $0$0 and $2\pi$2π both included because the interval is closed.

Step 5 — solve $\cos x = -\tfrac{1}{2}$cosx=−21​ on $[0, 2\pi]$[0,2π].

The principal value is $x = \arccos(-\tfrac{1}{2}) = \tfrac{2\pi}{3}$x=arccos(−21​)=32π​ (in the second quadrant). Cosine is also negative in the third quadrant; the symmetric partner is $x = 2\pi - \tfrac{2\pi}{3} = \tfrac{4\pi}{3}$x=2π−32π​=34π​.

M1 — correct principal value $\tfrac{2\pi}{3}$32π​.

A1 — correct second solution $\tfrac{4\pi}{3}$34π​ obtained by CAST or by symmetry of the cosine graph about $x = \pi$x=π.

Step 6 — assemble the full solution set.

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

A1 — all five solutions listed, in exact form, with no extraneous values.

B1 (presentation/AO2.5) — the candidate states the full solution set as a tidy list and confirms each value lies in $[0, 2\pi]$[0,2π]. Examiners reserve this final mark for solutions presented coherently rather than scattered through the working.

Total: 8 marks (M3 A4 B1).

---

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): Solve $2\sin^2\theta = 3\cos\theta$2sin2θ=3cosθ for $\theta \in [0°, 360°]$θ∈[0°,360°], giving answers correct to one decimal place where appropriate.

Mark scheme decomposition by AO:

• M1 (AO1.1a) — using the Pythagorean identity $\sin^2\theta = 1 - \cos^2\theta$sin2θ=1−cos2θ to convert the equation into a single trig function.
• A1 (AO1.1b) — obtaining the quadratic $2\cos^2\theta + 3\cos\theta - 2 = 0$2cos2θ+3cosθ−2=0.
• M1 (AO1.1b) — factorising as $(2\cos\theta - 1)(\cos\theta + 2) = 0$(2cosθ−1)(cosθ+2)=0.
• A1 (AO2.3) — rejecting $\cos\theta = -2$cosθ=−2 explicitly, with the reason "$\cos\theta \in [-1, 1]$cosθ∈[−1,1] for all real $\theta$θ".
• M1 (AO1.1b) — solving $\cos\theta = \tfrac{1}{2}$cosθ=21​: principal value $\theta = 60°$θ=60°.
• A1 (AO2.5) — the second solution $\theta = 360° - 60° = 300°$θ=360°−60°=300°, with both answers listed in the interval and presented to the requested accuracy.

Total: 6 marks split AO1 = 4, AO2 = 2. AO2 marks here are concentrated in (i) explicitly rejecting the impossible root with a stated reason and (ii) the symmetry argument that delivers the second solution. Both are reasoning marks, distinct from the procedural AO1 marks for substitution and factorisation.

---

Synoptic links

Connects to:

1. Identities (Pythagorean, compound-angle, double-angle): every non-trivial trig equation requires identity substitution before solving. Equations mixing $\sin^2\theta$sin2θ with $\cos\theta$cosθ collapse via $\sin^2\theta = 1 - \cos^2\theta$sin2θ=1−cos2θ; equations with $\sin 2x$sin2x alongside $\sin x$sinx or $\cos x$cosx collapse via $\sin 2x = 2\sin x \cos x$sin2x=2sinxcosx. The choice of substitution is the mark-bearing decision.

2. Calculus stationary points (section 9): "find the values of $x$x where $\frac{dy}{dx} = 0$dxdy​=0" with $y = \sin x + \cos 2x$y=sinx+cos2x produces $\cos x - 2\sin 2x = 0$cosx−2sin2x=0, which after $\sin 2x = 2\sin x \cos x$sin2x=2sinxcosx becomes $\cos x(1 - 4\sin x) = 0$cosx(1−4sinx)=0. The trig-equation skill is the calculus answer.

3. Harmonic form combination: equations of shape $a\sin\theta + b\cos\theta = c$asinθ+bcosθ=c are solved by first writing the LHS as $R\sin(\theta + \alpha)$Rsin(θ+α) with $R = \sqrt{a^2 + b^2}$R=a2+b2​. The problem then reduces to $\sin(\theta + \alpha) = c/R$sin(θ+α)=c/R — a single transformed-argument equation with the standard CAST treatment.

4. Complex roots (further-maths and university): the equation $e^{i\theta} = 1$eiθ=1 has solutions $\theta = 2n\pi$θ=2nπ for $n \in \mathbb{Z}$n∈Z. More generally $e^{i\theta} = z$eiθ=z for $|z| = 1$∣z∣=1 unwraps to $\cos\theta = \Re(z)$cosθ=ℜ(z) and $\sin\theta = \Im(z)$sinθ=ℑ(z), the same trig-equation-solving moves but viewed through Euler's identity.

5. Mechanics SHM (Paper 3): a particle's displacement $x = A\cos(\omega t + \phi)$x=Acos(ωt+ϕ) first reaches displacement $x_0$x0​ at time $t$t satisfying $\cos(\omega t + \phi) = x_0/A$cos(ωt+ϕ)=x0​/A. Every "when does the particle..." question is a transformed-argument trig equation, with the additional requirement that $t \geq 0$t≥0 (so principal-value selection differs from the symmetric $[0, 2\pi]$[0,2π] case).

---

Mark-scheme literacy

Solving-equation questions on 9MA0 Paper 2 typically split AO weight as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Applying the right identity, factorising, solving the principal equation, listing solutions in interval
AO2 (reasoning / interpretation)	25–35%	Justifying why a candidate root is rejected (out of range), using CAST/symmetry to find all solutions, presenting solutions in the requested form (degrees vs radians, exact vs decimal)
AO3 (problem-solving)	10–20%	Choosing which identity to apply when the question doesn't telegraph it; recognising a hidden quadratic; spotting when harmonic form is the entry route

Examiner-rewarded phrasing: "finding all solutions in $[0, 2\pi]$[0,2π]"; "factorising rather than dividing by trig terms (which would lose roots)"; "the principal value is $\arcsin(\tfrac{1}{2}) = \tfrac{\pi}{6}$arcsin(21​)=6π​, and by symmetry of the sine graph about $x = \tfrac{\pi}{2}$x=2π​ the second solution is $\pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}$π−6π​=65π​"; "rejecting $\cos\theta = -2$cosθ=−2 since cosine takes values in $[-1, 1]$[−1,1]"; "since the argument is $2x$2x and $x \in [0, 2\pi]$x∈[0,2π], the search range for $2x$2x is $[0, 4\pi]$[0,4π]".

Phrases that lose marks: "by my calculator, $x = 0.524$x=0.524" (a decimal where exact form is requested); "dividing both sides by $\sin\theta$sinθ" (lost roots); "$x = \arcsin(0.5) = 30°$x=arcsin(0.5)=30°" stopped there (only one solution in a multi-solution interval); "$x \approx 1.05$x≈1.05 rad" with no second solution given. Examiners explicitly flag these as method errors costing the A1 or the final A2.

A specific Edexcel pattern: when an interval is given in radians, answer in radians; when given in degrees, answer in degrees. Mixing the two is treated as a unit error on a Paper 2 question and forfeits the final accuracy mark.

---

Grade-band model answers

3-mark question

Question: Solve $\sin x = \tfrac{1}{2}$sinx=21​ for $x \in [0, 2\pi]$x∈[0,2π], giving exact answers.

Grade C response (~210 words):

The principal value is $x = \arcsin(\tfrac{1}{2}) = \tfrac{\pi}{6}$x=arcsin(21​)=6π​.

Sine is also positive in the second quadrant, so the second solution is $x = \pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}$x=π−6π​=65π​.

So $x = \tfrac{\pi}{6}$x=6π​ or $x = \tfrac{5\pi}{6}$x=65π​.