OCR C3 — Question 6 8 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeExpress and solve equation
DifficultyStandard +0.8 This is a two-part harmonic form question requiring standard R-cos(x-α) conversion followed by a non-trivial trigonometric equation. Part (i) is routine C3 content, but part (ii) requires recognizing that 6cos²x + sin2x can be rewritten using the double angle formula and cos²x = (1+cos2x)/2, then connecting it to part (i). The multi-step algebraic manipulation and the need to find all solutions in the given range elevate this above average difficulty, though it remains within standard C3 scope.
Spec1.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals

6. (i) Express \(3 \cos x ^ { \circ } + \sin x ^ { \circ }\) in the form \(R \cos ( x - \alpha ) ^ { \circ }\) where \(R > 0\) and \(0 < \alpha < 90\).
(ii) Using your answer to part (a), or otherwise, solve the equation $$6 \cos ^ { 2 } x ^ { \circ } + \sin 2 x ^ { \circ } = 0$$ for \(x\) in the interval \(0 \leq x \leq 360\), giving your answers to 1 decimal place where appropriate.

Question 6:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Notes
\(R\cos\alpha = 3\), \(R\sin\alpha = 1\)M1
\(R = \sqrt{3^2 + 1^2} = \sqrt{10}\)A1
\(\tan\alpha = \frac{1}{3}\), \(\alpha = 18.4\) (3sf)A1
\(\therefore 3\cos x° + \sin x° = \sqrt{10}\cos(x - 18.4)°\)
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Notes
\(6\cos^2 x + 2\sin x\cos x = 0\)M1
\(2\cos x(3\cos x + \sin x) = 0\)M1
\(\cos x = 0\) or \(3\cos x + \sin x = \sqrt{10}\cos(x - 18.4) = 0\)A1
\(x = 90, 270\) or \(x - 18.4 = 90, 270\)
\(x = 90, 108.4\) (1dp), \(270, 288.4\) (1dp)A2 (8)
# Question 6:

## Part (i):
| Answer/Working | Mark | Notes |
|---|---|---|
| $R\cos\alpha = 3$, $R\sin\alpha = 1$ | M1 | |
| $R = \sqrt{3^2 + 1^2} = \sqrt{10}$ | A1 | |
| $\tan\alpha = \frac{1}{3}$, $\alpha = 18.4$ (3sf) | A1 | |
| $\therefore 3\cos x° + \sin x° = \sqrt{10}\cos(x - 18.4)°$ | | |

## Part (ii):
| Answer/Working | Mark | Notes |
|---|---|---|
| $6\cos^2 x + 2\sin x\cos x = 0$ | M1 | |
| $2\cos x(3\cos x + \sin x) = 0$ | M1 | |
| $\cos x = 0$ or $3\cos x + \sin x = \sqrt{10}\cos(x - 18.4) = 0$ | A1 | |
| $x = 90, 270$ or $x - 18.4 = 90, 270$ | | |
| $x = 90, 108.4$ (1dp), $270, 288.4$ (1dp) | A2 | **(8)** |

---
6. (i) Express $3 \cos x ^ { \circ } + \sin x ^ { \circ }$ in the form $R \cos ( x - \alpha ) ^ { \circ }$ where $R > 0$ and $0 < \alpha < 90$.\\
(ii) Using your answer to part (a), or otherwise, solve the equation

$$6 \cos ^ { 2 } x ^ { \circ } + \sin 2 x ^ { \circ } = 0$$

for $x$ in the interval $0 \leq x \leq 360$, giving your answers to 1 decimal place where appropriate.\\

\hfill \mbox{\textit{OCR C3  Q6 [8]}}