What is 5.7: Trig equations?

Solve first for the trigonometric ratio, use a reference angle and the signs in each quadrant, then list only solutions in the stated interval.

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.

Explanation

Solve first for the trigonometric ratio, use a reference angle and the signs in each quadrant, then list only solutions in the stated interval.
For a quadratic in one trigonometric function, substitute a temporary variable, factorise or use the quadratic formula, and reject any ratio outside its possible range before solving each remaining branch.
When the equation involves $kx$ , transform the given interval for $x$ into the corresponding interval for $kx$ , find every solution there, and divide only at the end.
Inverse-trigonometric buttons return a principal value rather than the full solution set; endpoints, excluded endpoints and degree-versus-radian mode must all be checked explicitly.

Worked example

Determine all $x$ satisfying $2\cos^2x-3\cos x+1=0$ for $0\leq x<2\pi$ .

1.Factorise to $(2\cos x-1)(\cos x-1)=0$ .
2.Hence $\cos x=1/2$ or $\cos x=1$ .
3.In the interval, $\cos x=1/2$ at $x=\pi/3$ and $5\pi/3$ , while $\cos x=1$ at $x=0$ ; $2\pi$ is excluded.

Answer: $x=0,\ \pi/3,\ 5\pi/3$

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.

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)

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.

Other points in 5 Trigonometry

Want help turning this into marks?

Bring 5.7 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call