Question 8 - A-Level Maths

Edexcel C2 2009 January — Question 8 8 marks

Exam Board	Edexcel
Module	C2 (Core Mathematics 2)
Year	2009
Session	January
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Quadratic trigonometric equations
Type	Show then solve: sin²/cos² substitution
Difficulty	Moderate -0.3 This is a standard C2 trigonometric equation requiring the identity sin²x + cos²x = 1 to convert to a quadratic in cos x, then solving using the quadratic formula. The technique is routine and well-practiced, though the extended range to 720° adds a minor complication. Slightly easier than average due to the straightforward substitution and clear structure.
Spec	1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 1.05o Trigonometric equations: solve in given intervals

8. (a) Show that the equation $$4 \sin ^ { 2 } x + 9 \cos x - 6 = 0$$ can be written as $$4 \cos ^ { 2 } x - 9 \cos x + 2 = 0$$ (b) Hence solve, for $0 \leqslant x < 720 ^ { \circ }$, $$4 \sin ^ { 2 } x + 9 \cos x - 6 = 0$$ giving your answers to 1 decimal place.

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Question 8:

Part (a):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$4(1 - \cos^2 x) + 9\cos x - 6 = 0 \Rightarrow 4\cos^2 x - 9\cos x + 2 = 0$	M1 A1	Uses $\sin^2 x = 1 - \cos^2 x$; must have $= 0$

Part (b):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$(4\cos x - 1)(\cos x - 2) = 0 \Rightarrow \cos x = \ldots, \frac{1}{4}$	M1 A1	Solves quadratic; obtains $\frac{1}{4}$ accurately
$x = 75.5°$ $(\alpha)$	B1	Allow answers rounding to 75.5
$360 - \alpha$, $360 + \alpha$ or $720 - \alpha$	M1, M1	ft their value of $\alpha$
$284.5, \ 435.5, \ 644.5$	A1	Three and only three correct exact answers

# Question 8:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $4(1 - \cos^2 x) + 9\cos x - 6 = 0 \Rightarrow 4\cos^2 x - 9\cos x + 2 = 0$ | M1 A1 | Uses $\sin^2 x = 1 - \cos^2 x$; must have $= 0$ |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(4\cos x - 1)(\cos x - 2) = 0 \Rightarrow \cos x = \ldots, \frac{1}{4}$ | M1 A1 | Solves quadratic; obtains $\frac{1}{4}$ accurately |
| $x = 75.5°$ $(\alpha)$ | B1 | Allow answers rounding to 75.5 |
| $360 - \alpha$, $360 + \alpha$ or $720 - \alpha$ | M1, M1 | ft their value of $\alpha$ |
| $284.5, \ 435.5, \ 644.5$ | A1 | Three **and only three** correct exact answers |

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8. (a) Show that the equation

$$4 \sin ^ { 2 } x + 9 \cos x - 6 = 0$$

can be written as

$$4 \cos ^ { 2 } x - 9 \cos x + 2 = 0$$

(b) Hence solve, for $0 \leqslant x < 720 ^ { \circ }$,

$$4 \sin ^ { 2 } x + 9 \cos x - 6 = 0$$

giving your answers to 1 decimal place.

\hfill \mbox{\textit{Edexcel C2 2009 Q8 [8]}}

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