What is 5.7: Trig equations?

Rearrange to a single ratio, find the principal value from the calculator, then use the CAST diagram or the graph to generate every solution in the interval: for cos the second solution is 360°-θ or 2π-θ, for sin it is 180°-θ, for tan add 180°.

How many marks are 5.7 questions worth?

5.7 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 5.7 appeared in?

We have not yet indexed a verified past-paper appearance for 5.7 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

5.7

Solve simple trigonometric equations in a given interval, including quadratic equations in sin, cos and tan and equations involving multiples of the unknown angle.

Draft — not yet indexed

Trig equations

Worked answers and methods for 5.7 on Edexcel A-level Maths 9MA0.

Definition

Specification 5.7: solve simple trigonometric equations in a given interval, including quadratic equations in $\sin$ , $\cos$ and $\tan$ , and equations involving multiples of the unknown angle.

Explanation

Rearrange to a single ratio, find the principal value from the calculator, then use the CAST diagram or the graph to generate every solution in the interval: for $\cos$ the second solution is $360^\circ-\theta$ or $2\pi-\theta$ , for $\sin$ it is $180^\circ-\theta$ , for $\tan$ add $180^\circ$ .
A quadratic in one ratio is factorised or solved by the formula, and any root outside $[-1,1]$ for $\sin$ or $\cos$ is rejected before solving each branch; a Pythagorean identity converts a mixed equation into a quadratic in one ratio first.
For a multiple angle, transform the interval, so $0<x<90^\circ$ becomes $0<2x<180^\circ$ , find every value of $2x$ in that range, and divide at the end.
Never divide by $\sin\theta$ or $\cos\theta$ , which loses solutions; factorise instead.
Check the mode, degrees or radians, against the interval, and check endpoints against strict or non-strict inequalities.

Worked example

Solve, for

0\leq x<2\pi

, the equation

2\cos^2x-3\cos x+1=0

(4)

(Total for Question 1 is 4 marks)

Working

1.Factorise as a quadratic in $\cos x$ : $(2\cos x-1)(\cos x-1)=0$ .
2.So $\cos x=\dfrac12$ or $\cos x=1$ .
3. $\cos x=\dfrac12$ gives $x=\dfrac{\pi}{3}$ and $x=2\pi-\dfrac{\pi}{3}=\dfrac{5\pi}{3}$ ; $\cos x=1$ gives $x=0$ only, since $2\pi$ is excluded.

Answer: $x=0,\ \dfrac{\pi}{3},\ \dfrac{5\pi}{3}$

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Attempts to solve the three-term quadratic in $\cos x$ , eg $(2\cos x-1)(\cos x-1)$	M1
	$\cos x=\dfrac12$ and $\cos x=1$	A1
	Obtains $x=\dfrac{\pi}{3}$ and a second solution from $2\pi-$ their principal value	M1
	$x=0,\ \dfrac{\pi}{3},\ \dfrac{5\pi}{3}$ and no extra solutions in the interval	A1
Notes
A1: Degrees score A0

(4 marks)

Common mistakes

Don't give only the principal value returned by the inverse-trigonometric button.
Don't include an excluded endpoint or miss solutions created by the multiple angle.

Mark-scheme wording to use

“M1: Correct order of operations from tan² 2x = α or sin² 2x = β or cos² 2x = γ or cos 4x = k or equivalents e.g. cosec² 2x = 1/β where α > 0 or 0 < β < 1 or 0 < γ < 1 or 0 < k < 1 leading to at least one value for x e.g. square roots, finds inverse tan/sin/cos/cosec and divides by 2 or inverse cos and divides by 4 … A1: Correct values. Allow awrt 17.6° and awrt 72.4°. The degrees symbol is not required. … if there are extra angles in range score A0. Answers in radians score A0.” — Pearson 9MA0, June 2024 Paper 2, Q8(b)
“M1 for using cos^2(x) = 1 - sin^2(x) to form a quadratic in sin(x). A1 for 12sin^2(x) + sin(x) - 1 = 0. M1 for attempting to solve that quadratic. A1 for sin(x) = 1/4 and -1/3. M1 for inverse-trigonometric work producing two correct values. A1 for all four values 14.48, 165.52, -19.47 and -160.53 degrees” — Pearson 9MA0, Sample Paper 2, Q12(a)
“B1 for explaining that squaring introduced an extraneous solution or lost the sign information” — Pearson 9MA0, Sample Paper 2, Q2(b)(ii)

What the examiner wants

Every in-range solution and nothing else. The June 2024 Paper 2 scheme withholds the final mark for radians in a degrees question or for extra in-range angles, and the Paper 1 report on the trigonometric model praises candidates who identified that degrees, not radians, were required before solving.

Exam tip

Solve for the trig value first, then use symmetry and periodicity to list every solution in the stated interval.

Worked practice

Tier 1 · Easy

Solve

\sin\theta=0.4

for

0\leq\theta\leq2\pi

, giving solutions to

3

decimal places.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\theta=0.412$ or $2.730$	3
Notes
The principal value is $\arcsin(0.4)=0.4115\ldots$ . Sine is positive in quadrants I and II, so the second solution is $\pi-0.4115\ldots=2.7301\ldots$ . Thus $\theta=0.412$ or $2.730$ .

(3 marks)

Tier 2 · Standard

Solve

\sin(2x)=\dfrac{\sqrt3}{2}

for

0\leq x\leq2\pi

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$x=\dfrac{\pi}{6},\ \dfrac{\pi}{3},\ \dfrac{7\pi}{6},\ \dfrac{4\pi}{3}$	4
Notes
Since $0\leq2x\leq4\pi$ , solve over two complete periods. The solutions for $2x$ are $\pi/3$ , $2\pi/3$ , $7\pi/3$ and $8\pi/3$ . Dividing each by $2$ gives $x=\pi/6,\pi/3,7\pi/6,4\pi/3$ .

(4 marks)

Tier 3 · Hard

Find every solution of

\tan(2x)=-\sqrt3

in the interval

-\pi/2\leq x\leq\pi

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$x=-\pi/6,\ \pi/3,\ 5\pi/6$	5
Notes
The transformed interval is $-\pi\leq2x\leq2\pi$ . Since tangent has period $\pi$ and reference angle $\pi/3$ , the solutions for $2x$ in this interval are $-\pi/3$ , $2\pi/3$ and $5\pi/3$ . Dividing each by $2$ gives $x=-\pi/6$ , $\pi/3$ and $5\pi/6$ .

(5 marks)

Tier 1 · Easy

Solve

\cos x=-\dfrac{\sqrt2}{2}

for

0\leq x<2\pi

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$x=\dfrac{3\pi}{4},\ \dfrac{5\pi}{4}$	2
Notes
The reference angle is $\pi/4$ . Cosine is negative in quadrants II and III, giving $x=3\pi/4$ and $x=5\pi/4$ .

(2 marks)

Tier 2 · Standard

Solve

2\sin^2x+\sin x-1=0

for

0\leq x<2\pi

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6},\ \dfrac{3\pi}{2}$	4
Notes
Factorise to $(2\sin x-1)(\sin x+1)=0$ . Thus $\sin x=1/2$ or $\sin x=-1$ . In the stated interval these give $x=\pi/6,5\pi/6$ and $3\pi/2$ respectively.

(4 marks)

Tier 3 · Hard

Solve

4\sin^2(3x)-1=0

for

-\pi/2\leq x\leq2\pi/3

. Show the complete set of values of

3x

in the corresponding interval.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$3x=-\dfrac{7\pi}{6},\ -\dfrac{5\pi}{6},\ -\dfrac{\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6},\ \dfrac{7\pi}{6},\ \dfrac{11\pi}{6}$ (for $-3\pi/2\leq3x\leq2\pi$ ) $x=-\dfrac{7\pi}{18},\ -\dfrac{5\pi}{18},\ -\dfrac{\pi}{18},\ \dfrac{\pi}{18},\ \dfrac{5\pi}{18},\ \dfrac{7\pi}{18},\ \dfrac{11\pi}{18}$	6
Notes
Let $y=3x$ , so $-3\pi/2\leq y\leq2\pi$ . Factorising gives $(2\sin y-1)(2\sin y+1)=0$ , so $\sin y=1/2$ or $\sin y=-1/2$ . For $\sin y=1/2$ , the values in the expanded interval are $y=-7\pi/6,\pi/6,5\pi/6$ . For $\sin y=-1/2$ , they are $y=-5\pi/6,-\pi/6,7\pi/6,11\pi/6$ . Dividing all seven values by $3$ and ordering them gives $x=-7\pi/18,-5\pi/18,-\pi/18,\pi/18,5\pi/18,7\pi/18,11\pi/18$ .

(6 marks)

Tier 2 · Standard

Solve

2\sin x\cos x=\sin x

for

0\leq x<2\pi

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$x=0,\ \pi/3,\ \pi,\ 5\pi/3$	4
Notes
Bring all terms to one side and factor without dividing by $\sin x$ : $\sin x(2\cos x-1)=0$ . Thus $\sin x=0$ or $\cos x=1/2$ . In $0\leq x<2\pi$ , the first branch gives $x=0,\pi$ and the second gives $x=\pi/3,5\pi/3$ . These four values form the complete solution set.

(4 marks)

Tier 3 · Hard

Determine the positive constant

R

and acute angle

\alpha

such that

3\sin x+4\cos x=R\sin(x+\alpha)

. Hence solve

3\sin x+4\cos x=2

for

-\pi\leq x\leq\pi

, giving solutions to

3

decimal places. Use unrounded values in your working.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$3\sin x+4\cos x=5\sin(x+\alpha)$ , where $\alpha=\arctan(4/3)$ $x=-0.516,\ 1.803$	5
Notes
Expanding $R\sin(x+\alpha)$ gives $R\cos\alpha=3$ and $R\sin\alpha=4$ , so $R=5$ and $\alpha=\arctan(4/3)$ . The equation becomes $\sin(x+\alpha)=2/5$ . Let $\beta=\arcsin(2/5)$ . Since $-\pi+\alpha\leq x+\alpha\leq\pi+\alpha$ , the only values in the transformed interval are $x+\alpha=\beta$ and $x+\alpha=\pi-\beta$ . Hence $x=\beta-\alpha=-0.515778\ldots$ or $x=\pi-\beta-\alpha=1.802780\ldots$ , giving $x=-0.516,1.803$ .

(5 marks)

Tier 3 · Hard

Solve

\cos(2x)=2\sin x

for

0\leq x<2\pi

, giving all solutions exactly.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$x=\arcsin\left(\dfrac{\sqrt3-1}{2}\right),\ \pi-\arcsin\left(\dfrac{\sqrt3-1}{2}\right)$	5
Notes
Use $\cos(2x)=1-2\sin^2x$ . With $s=\sin x$ , the equation becomes $1-2s^2=2s$ , or $2s^2+2s-1=0$ . Thus $s=(-1\pm\sqrt3)/2$ . The value $(-1-\sqrt3)/2$ is less than $-1$ and is rejected, leaving $\sin x=(\sqrt3-1)/2$ . This value is positive, so the complete solution set in the interval consists of the quadrant-I value and its quadrant-II partner, as stated.

(5 marks)

Q10

Tier 3 · Hard

10.

Determine all real values of

k

for which

\sin^2x-\sin x=k

has exactly three distinct solutions in the interval

0\leq x<2\pi

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$k=0$	6
Notes
Let $s=\sin x$ , so $-1\leq s\leq1$ and $s^2-s=k$ . Each root strictly between $-1$ and $1$ produces two values of $x$ in the interval, while either endpoint $s=1$ or $s=-1$ produces one value. A total of three solutions therefore requires one endpoint root and one interior root. If $s=1$ , then $k=0$ and the other root is $s=0$ , giving one solution from $s=1$ and two from $s=0$ . If $s=-1$ , then $k=2$ and the other root is $s=2$ , which is outside the sine range, so there is only one solution. A repeated interior root produces two solutions, and two interior roots produce four. Hence only $k=0$ gives exactly three distinct solutions.

(6 marks)

Q11

Tier 2 · Standard

11.

Solve, for

0^\circ\leq x<360^\circ

, the equation

2\cos^2x+3\sin x=3

(5)

(Total for Question 11 is 5 marks)

Mark scheme

Mark scheme for question 11
Question	Scheme	Marks
11	$x=30^\circ,\ 90^\circ,\ 150^\circ$	5
Notes
Replace $\cos^2x$ by $1-\sin^2x$ : $2-2\sin^2x+3\sin x=3$ , so $2\sin^2x-3\sin x+1=0$ . This factorises as $(2\sin x-1)(\sin x-1)=0$ , so $\sin x=\tfrac12$ or $\sin x=1$ . In the interval, $\sin x=\tfrac12$ gives $x=30^\circ$ and $180^\circ-30^\circ=150^\circ$ , and $\sin x=1$ gives $x=90^\circ$ .

(5 marks)

Q12

Tier 2 · Standard

12.

Solve, for

0^\circ\leq\theta<180^\circ

, the equation

\tan(2\theta-30^\circ)=\sqrt3

(4)

(Total for Question 12 is 4 marks)

Mark scheme

Mark scheme for question 12
Question	Scheme	Marks
12	$\theta=45^\circ,\ 135^\circ$	4
Notes
For $0^\circ\leq\theta<180^\circ$ , the angle $2\theta-30^\circ$ lies in $-30^\circ\leq2\theta-30^\circ<330^\circ$ . The principal value is $\tan^{-1}\sqrt3=60^\circ$ , and tangent has period $180^\circ$ , so $2\theta-30^\circ=60^\circ$ or $240^\circ$ (the values $-120^\circ$ and $420^\circ$ lie outside the range). Hence $2\theta=90^\circ$ or $270^\circ$ , giving $\theta=45^\circ$ or $135^\circ$ .

(4 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for 5.7. Browse the Edexcel A-level Maths 9MA0 past papers directly.

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