OCR MEI C4 — Question 1 7 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTrigonometric equations in context
TypeSolve equation using double angle identity
DifficultyModerate -0.3 This is a straightforward two-part question requiring the standard double angle identity cos 2θ = 1 - 2sin²θ, followed by solving a quadratic equation in sin θ. While it involves multiple steps, both the identity substitution and quadratic solving are routine techniques that students practice extensively, making it slightly easier than average.
Spec1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals
1 Express \(6 \cos 2 \theta + \sin \theta\) in terms of \(\sin \theta\).
Hence solve the equation \(6 \cos 2 \theta + \sin \theta = 0\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
Question 1
M1* Use of \(\cos^2\theta = 1 - 2\sin^2\theta\) (maybe implied in substitution)
A1 \(6\cos^2\theta + \sin\theta = 0\) or \(12\sin^2\theta - \sin\theta - 6 = 0\)
M1dep* Use of correct quadratic equation formula or factorising or completing the square on their three term quadratic in \(\sin\theta\) provided \(b^2 - 4ac \geq 0\)
A1 \((4\sin\theta - 3)(3\sin\theta + 2) = 0\)
B1 \(\sin\theta = \frac{3}{4}\) or \(\sin\theta = -\frac{2}{3}\)
B1 \(\sin\theta = \frac{3}{4}\), \(\theta = 48.6°, 131.4°\)
B1 \(\sin\theta = -\frac{2}{3}\), \(\theta = 221.8°, 318.2°\)
Guidance:
- First correct solution to 1 d.p. or better (e.g. \(48.59°\) etc)
- Three correct solutions
- All four correct solutions and no others in the range
- Ignore solutions outside the range
- SC Award max B1B1B0 for answers in radians (\(0.85, 2.29, 3.87, 5.55\) or better – so one correct B1, three correct B1). Award max B1 if there are extra solutions in the range with radians
- SC If M1M1 awarded and both values of \(\sin\theta\) obtained but B0B0B0 then award B1 only for evidence of using \(\sin\theta = \sin(180° - \theta)\)
# Question 1

M1* Use of $\cos^2\theta = 1 - 2\sin^2\theta$ (maybe implied in substitution)

A1 $6\cos^2\theta + \sin\theta = 0$ or $12\sin^2\theta - \sin\theta - 6 = 0$

M1dep* Use of correct quadratic equation formula or factorising or completing the square on their three term quadratic in $\sin\theta$ provided $b^2 - 4ac \geq 0$

A1 $(4\sin\theta - 3)(3\sin\theta + 2) = 0$

B1 $\sin\theta = \frac{3}{4}$ or $\sin\theta = -\frac{2}{3}$

B1 $\sin\theta = \frac{3}{4}$, $\theta = 48.6°, 131.4°$

B1 $\sin\theta = -\frac{2}{3}$, $\theta = 221.8°, 318.2°$

**Guidance:**
- First correct solution to 1 d.p. or better (e.g. $48.59°$ etc)
- Three correct solutions
- All four correct solutions and no others in the range
- Ignore solutions outside the range
- SC Award max B1B1B0 for answers in radians ($0.85, 2.29, 3.87, 5.55$ or better – so one correct B1, three correct B1). Award max B1 if there are extra solutions in the range with radians
- SC If M1M1 awarded and both values of $\sin\theta$ obtained but B0B0B0 then award B1 only for evidence of using $\sin\theta = \sin(180° - \theta)$
1 Express $6 \cos 2 \theta + \sin \theta$ in terms of $\sin \theta$.\\
Hence solve the equation $6 \cos 2 \theta + \sin \theta = 0$, for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.

\hfill \mbox{\textit{OCR MEI C4  Q1 [7]}}
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