Question 2 - A-Level Maths

WJEC Unit 1 Specimen — Question 2 6 marks

Exam Board	WJEC
Module	Unit 1 (Unit 1)
Session	Specimen
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Standard trigonometric equations
Type	Convert to quadratic in sin/cos
Difficulty	Standard +0.3 This is a standard trigonometric equation requiring the identity sin²θ + cos²θ = 1 to convert to a single function, then solving a quadratic in sin θ. The multi-step process and finding all solutions in the given range makes it slightly above average, but it follows a well-practiced technique with no novel insight required.
Spec	1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 1.05o Trigonometric equations: solve in given intervals

Find all values of $\theta$ between $0°$ and $360°$ satisfying $$7 \sin^2 \theta + 1 = 3 \cos^2 \theta - \sin \theta.$$ [6]

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Answer	Marks	Guidance
$7 \sin^2 \theta + 3 = 3(1 – \sin^2 \theta) – \sin \theta$	M1	AO1 (correct use of $\cos^2 \theta = 1 – \sin^2 \theta$)
An attempt to collect terms, form and solve a quadratic equation in $\sin \theta$, either by using the quadratic formula or by getting the expression into the form $(a \sin \theta + b)(c \sin \theta + d)$, with $a \times c$ = candidate's coefficient of $\sin^2 \theta$ and $b \times d$ = candidate's constant	m1	AO1
$10 \sin^2 \theta + \sin \theta – 2 = 0$	—	—
$\Rightarrow (2 \sin \theta + 1)(5 \sin \theta – 2) = 0$	A1	AO1 (c.a.o.)
$\Rightarrow \sin \theta = -\frac{1}{2}, \sin \theta = \frac{2}{5}$	A1	AO1
$\theta = 210°, 330°$	B1	AO1
$\theta = 23.57(8178...)°, 156.42(182...)°$	B1	AO1

Note: Subtract 1 mark for each additional root in range for each branch, ignore roots outside range.

Answer	Marks	Guidance
$\sin \theta = +, -$, f.t. for 3 marks; $\sin \theta = -, -$, f.t. for 2 marks; $\sin \theta = +, +$, f.t. for 1 mark	—	—

Total: [6]

$7 \sin^2 \theta + 3 = 3(1 – \sin^2 \theta) – \sin \theta$ | M1 | AO1 (correct use of $\cos^2 \theta = 1 – \sin^2 \theta$)

An attempt to collect terms, form and solve a quadratic equation in $\sin \theta$, either by using the quadratic formula or by getting the expression into the form $(a \sin \theta + b)(c \sin \theta + d)$, with $a \times c$ = candidate's coefficient of $\sin^2 \theta$ and $b \times d$ = candidate's constant | m1 | AO1

$10 \sin^2 \theta + \sin \theta – 2 = 0$ | — | —
$\Rightarrow (2 \sin \theta + 1)(5 \sin \theta – 2) = 0$ | A1 | AO1 (c.a.o.)
$\Rightarrow \sin \theta = -\frac{1}{2}, \sin \theta = \frac{2}{5}$ | A1 | AO1

$\theta = 210°, 330°$ | B1 | AO1
$\theta = 23.57(8178...)°, 156.42(182...)°$ | B1 | AO1

Note: Subtract 1 mark for each additional root in range for each branch, ignore roots outside range.

$\sin \theta = +, -$, f.t. for 3 marks; $\sin \theta = -, -$, f.t. for 2 marks; $\sin \theta = +, +$, f.t. for 1 mark | — | —

**Total: [6]**

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Find all values of $\theta$ between $0°$ and $360°$ satisfying
$$7 \sin^2 \theta + 1 = 3 \cos^2 \theta - \sin \theta.$$ [6]

\hfill \mbox{\textit{WJEC Unit 1  Q2 [6]}}

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Q1 7 Q2 6 Q3 6 Q4 5 Q5 12 Q6 5 Q7 5 Q8 6 Q9 7 Q10 8 Q11 3 Q12 3 Q13 7 Q14 8 Q15 8 Q16 5 Q17 12 Q18 7

Answer	Marks	Guidance
\(7 \sin^2 \theta + 3 = 3(1 – \sin^2 \theta) – \sin \theta\)	M1	AO1 (correct use of \(\cos^2 \theta = 1 – \sin^2 \theta\))
An attempt to collect terms, form and solve a quadratic equation in \(\sin \theta\), either by using the quadratic formula or by getting the expression into the form \((a \sin \theta + b)(c \sin \theta + d)\), with \(a \times c\) = candidate's coefficient of \(\sin^2 \theta\) and \(b \times d\) = candidate's constant	m1	AO1
\(10 \sin^2 \theta + \sin \theta – 2 = 0\)	—	—
\(\Rightarrow (2 \sin \theta + 1)(5 \sin \theta – 2) = 0\)	A1	AO1 (c.a.o.)
\(\Rightarrow \sin \theta = -\frac{1}{2}, \sin \theta = \frac{2}{5}\)	A1	AO1
\(\theta = 210°, 330°\)	B1	AO1
\(\theta = 23.57(8178...)°, 156.42(182...)°\)	B1	AO1