Edexcel C2 2014 January — Question 9 9 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Year2014
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicQuadratic trigonometric equations
TypeShow then solve: compound angle substitution
DifficultyStandard +0.3 This is a standard C2 trigonometric equation requiring routine application of sin²x + cos²x = 1 to convert to quadratic form, then solving a quadratic and finding angles in a given range. Part (a) is algebraic manipulation with clear guidance, part (b) applies the same method with a substitution (2θ for x). Slightly above average due to the two-part structure and need to handle the double angle carefully, but follows a well-practiced technique with no novel insight required.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals

9. (a) Show that the equation $$5 \sin x - \cos ^ { 2 } x + 2 \sin ^ { 2 } x = 1$$ can be written in the form $$3 \sin ^ { 2 } x + 5 \sin x - 2 = 0$$ (b) Hence solve, for \(- 180 ^ { \circ } \leqslant \theta < 180 ^ { \circ }\), the equation $$5 \sin 2 \theta - \cos ^ { 2 } 2 \theta + 2 \sin ^ { 2 } 2 \theta = 1$$ giving your answers to 2 decimal places.

Question 9(a):
AnswerMarks Guidance
\(5\sin x - (1-\sin^2 x) + 2\sin^2 x = 1\)M1 Use of \(\cos^2 x = 1 - \sin^2 x\)
\(3\sin^2 x + 5\sin x - 2 = 0\)A1 Correct completion to printed answer
Question 9(b):
AnswerMarks Guidance
\((3\sin 2\theta - 1)(\sin 2\theta + 2) = 0 \Rightarrow \sin 2\theta = \ldots\)M1 Attempt to solve for \(\sin 2\theta\) or \(\sin\theta\)
\(\sin(2\theta)/\sin\theta = \frac{1}{3}\) (or \(-2\))A1
\(2\theta/\theta = \sin^{-1}\left(\frac{1}{3}\right)\)M1
\(2\theta = 19.47122\ldots\), \(\theta = 9.74\)A1 Awrt
\(2\theta/\theta = 180-19.47,\ -180-19.47,\ -360+19.47\ldots\)M1 At least one of these
\(\theta = 80.26,\ -99.74,\ -170.26\) (allow awrt \(80.3,\ -99.7,\ -170.3\))A1, A1 First A1: any two correct to awrt accuracy; Second A1: all values correct, no extra values in range
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## Question 9(a):

| $5\sin x - (1-\sin^2 x) + 2\sin^2 x = 1$ | M1 | Use of $\cos^2 x = 1 - \sin^2 x$ |
|---|---|---|
| $3\sin^2 x + 5\sin x - 2 = 0$ | A1 | Correct completion to printed answer |

## Question 9(b):

| $(3\sin 2\theta - 1)(\sin 2\theta + 2) = 0 \Rightarrow \sin 2\theta = \ldots$ | M1 | Attempt to solve for $\sin 2\theta$ or $\sin\theta$ |
|---|---|---|
| $\sin(2\theta)/\sin\theta = \frac{1}{3}$ (or $-2$) | A1 | — |
| $2\theta/\theta = \sin^{-1}\left(\frac{1}{3}\right)$ | M1 | — |
| $2\theta = 19.47122\ldots$, $\theta = 9.74$ | A1 | Awrt |
| $2\theta/\theta = 180-19.47,\ -180-19.47,\ -360+19.47\ldots$ | M1 | At least one of these |
| $\theta = 80.26,\ -99.74,\ -170.26$ (allow awrt $80.3,\ -99.7,\ -170.3$) | A1, A1 | First A1: any two correct to awrt accuracy; Second A1: all values correct, no extra values in range |

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Could you please share the actual mark scheme pages with the question content, answers, and mark allocations? I'd be happy to format those for you once I can see the content.
9. (a) Show that the equation

$$5 \sin x - \cos ^ { 2 } x + 2 \sin ^ { 2 } x = 1$$

can be written in the form

$$3 \sin ^ { 2 } x + 5 \sin x - 2 = 0$$

(b) Hence solve, for $- 180 ^ { \circ } \leqslant \theta < 180 ^ { \circ }$, the equation

$$5 \sin 2 \theta - \cos ^ { 2 } 2 \theta + 2 \sin ^ { 2 } 2 \theta = 1$$

giving your answers to 2 decimal places.\\

\hfill \mbox{\textit{Edexcel C2 2014 Q9 [9]}}