Question 4 - A-Level Maths

Edexcel C2 2005 January — Question 4 7 marks

Exam Board	Edexcel
Module	C2 (Core Mathematics 2)
Year	2005
Session	January
Marks	7
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 Pythagorean identity (cos²x = 1 - sin²x) to convert to a quadratic, then factorization or quadratic formula, followed by routine inverse trig. The conversion is shown in part (a), making part (b) straightforward. Slightly easier than average due to the scaffolding and standard technique.
Spec	1.02f Solve quadratic equations: including in a function of unknown 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 1.05o Trigonometric equations: solve in given intervals

4. (a) Show that the equation $$5 \cos ^ { 2 } x = 3 ( 1 + \sin x )$$ can be written as $$5 \sin ^ { 2 } x + 3 \sin x - 2 = 0 .$$ (b) Hence solve, for $0 \leqslant x < 360 ^ { \circ }$, the equation $$5 \cos ^ { 2 } x = 3 ( 1 + \sin x )$$ giving your answers to 1 decimal place where appropriate.

Show mark scheme Show mark scheme source

(a) $5(1-\sin^2 x) = 3(1+\sin x)$

$5 - 5\sin^2 x = 3 + 3\sin x$

Answer	Marks	Guidance
$0 = 5\sin^2 x + 3\sin x - 2 *$	M1, A1 cso	(2 marks)

A1 cso: Final answer only.

(b) $0 = (5\sin x - 2)(\sin x + 1)$

Answer	Marks	Guidance
$\sin x = \frac{2}{5}, -1$	(both)	M1, A1, B1, M1, B1

1st B1: Correct solution, $\alpha$ to $\sin x = \frac{2}{5}$. Must be 1 d.p.

2nd M1: $180 - \alpha$, accept nearest degree or awrt.

Answer only in (b) scores M0A0 but then could score B1M1B1.

Incorrect factorisation probably only gets $\frac{2}{5}$.

Answer	Marks
$\sin x = \frac{2}{5} \Rightarrow x = 23.6$	($\alpha = 23.6$ or $156.4$)
$, 156.4$	(180- $\alpha$)
$\sin x = -1 \Rightarrow x = 270$

(Subtotal: 7 marks)

**(a)** $5(1-\sin^2 x) = 3(1+\sin x)$
$5 - 5\sin^2 x = 3 + 3\sin x$
$0 = 5\sin^2 x + 3\sin x - 2 *$ | M1, A1 cso | (2 marks) | M1: For use of $\cos^2 x = 1 - \sin^2 x$. Condone missing parentheses.
A1 cso: Final answer only.

**(b)** $0 = (5\sin x - 2)(\sin x + 1)$
$\sin x = \frac{2}{5}, -1$ | (both) | M1, A1, B1, M1, B1 | (5 marks) | 1st M1: Attempt to solve $\to \sin x =$
1st B1: Correct solution, $\alpha$ to $\sin x = \frac{2}{5}$. Must be 1 d.p.
2nd M1: $180 - \alpha$, accept nearest degree or awrt.
Answer only in (b) scores M0A0 but then could score B1M1B1.
Incorrect factorisation probably only gets $\frac{2}{5}$.

$\sin x = \frac{2}{5} \Rightarrow x = 23.6$ | ($\alpha = 23.6$ or $156.4$) | 
$, 156.4$ | (180- $\alpha$) |
$\sin x = -1 \Rightarrow x = 270$ |  |

**(Subtotal: 7 marks)**

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4. (a) Show that the equation

$$5 \cos ^ { 2 } x = 3 ( 1 + \sin x )$$

can be written as

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

(b) Hence solve, for $0 \leqslant x < 360 ^ { \circ }$, the equation

$$5 \cos ^ { 2 } x = 3 ( 1 + \sin x )$$

giving your answers to 1 decimal place where appropriate.\\

\hfill \mbox{\textit{Edexcel C2 2005 Q4 [7]}}

This paper (9 questions)

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Q1 4 Q2 6 Q3 7 Q4 7 Q5 8 Q6 8 Q7 11 Q8 12 Q9 12

Answer	Marks
\(\sin x = \frac{2}{5} \Rightarrow x = 23.6\)	(\(\alpha = 23.6\) or \(156.4\))
\(, 156.4\)	(180- \(\alpha\))
\(\sin x = -1 \Rightarrow x = 270\)

Edexcel C2 2005 January — Question 4 7 marks

(a) \(5(1-\sin^2 x) = 3(1+\sin x)\)

\(5 - 5\sin^2 x = 3 + 3\sin x\)

A1 cso: Final answer only.

(b) \(0 = (5\sin x - 2)(\sin x + 1)\)

1st B1: Correct solution, \(\alpha\) to \(\sin x = \frac{2}{5}\). Must be 1 d.p.

2nd M1: \(180 - \alpha\), accept nearest degree or awrt.

Answer only in (b) scores M0A0 but then could score B1M1B1.

Incorrect factorisation probably only gets \(\frac{2}{5}\).

(Subtotal: 7 marks)

This paper (9 questions)