Question 5 - A-Level Maths

Edexcel C3 — Question 5 10 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Harmonic Form
Type	Express and solve equation
Difficulty	Standard +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 (a) is routine C3 content, but part (b) requires recognizing that 6cos²x = 3(1+cos2x), converting to the harmonic form, and solving a quadratic-type equation. The multi-step manipulation and connection between parts elevates this above average difficulty.
Spec	1.05l Double angle formulae: and compound angle formulae 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc 1.05o Trigonometric equations: solve in given intervals

5. (a) Express $3 \cos x ^ { \circ } + \sin x ^ { \circ }$ in the form $R \cos ( x - \alpha ) ^ { \circ }$ where $R > 0$ and $0 < \alpha < 90$.
(b) 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.

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Answer	Marks	Guidance
(a) $3\cos x + \sin x = R\cos x \cos \alpha + R\sin x \sin \alpha$	M1 A1
$R\cos \alpha = 3$, $R\sin \alpha = 1$	M1 A1
$\therefore R = \sqrt{3^2 + 1^2} = \sqrt{10}$	M1 A1
$\tan \alpha = \frac{1}{3}$, $\alpha = 18.4$ (3sf)	M1 A1
$\therefore 3\cos x^° + \sin x^° = \sqrt{10}\cos(x - 18.4)^°$	M1 A1
(b) $6\cos^2 x + 2\sin x \cos x = 0$	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 $x - 18.4 = 90, 270$	B1 M1
$x = 90, 108.4$ (1dp), $270, 288.4$ (1dp)	A1	(10 marks)

**(a)** $3\cos x + \sin x = R\cos x \cos \alpha + R\sin x \sin \alpha$ | M1 A1 |

$R\cos \alpha = 3$, $R\sin \alpha = 1$ | M1 A1 |

$\therefore R = \sqrt{3^2 + 1^2} = \sqrt{10}$ | M1 A1 |

$\tan \alpha = \frac{1}{3}$, $\alpha = 18.4$ (3sf) | M1 A1 |

$\therefore 3\cos x^° + \sin x^° = \sqrt{10}\cos(x - 18.4)^°$ | M1 A1 |

**(b)** $6\cos^2 x + 2\sin x \cos x = 0$ | 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 $x - 18.4 = 90, 270$ | B1 M1 |

$x = 90, 108.4$ (1dp), $270, 288.4$ (1dp) | A1 | (10 marks)

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5. (a) Express $3 \cos x ^ { \circ } + \sin x ^ { \circ }$ in the form $R \cos ( x - \alpha ) ^ { \circ }$ where $R > 0$ and $0 < \alpha < 90$.\\
(b) 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{Edexcel C3  Q5 [10]}}

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Q1 4 Q2 6 Q3 8 Q4 10 Q5 10 Q6 12 Q7 12 Q8 13

Answer	Marks	Guidance
(a) \(3\cos x + \sin x = R\cos x \cos \alpha + R\sin x \sin \alpha\)	M1 A1
\(R\cos \alpha = 3\), \(R\sin \alpha = 1\)	M1 A1
\(\therefore R = \sqrt{3^2 + 1^2} = \sqrt{10}\)	M1 A1
\(\tan \alpha = \frac{1}{3}\), \(\alpha = 18.4\) (3sf)	M1 A1
\(\therefore 3\cos x^° + \sin x^° = \sqrt{10}\cos(x - 18.4)^°\)	M1 A1
(b) \(6\cos^2 x + 2\sin x \cos x = 0\)	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 \(x - 18.4 = 90, 270\)	B1 M1
\(x = 90, 108.4\) (1dp), \(270, 288.4\) (1dp)	A1	(10 marks)