Question 6 - A-Level Maths

Edexcel C2 2016 June — Question 6 9 marks

Exam Board	Edexcel
Module	C2 (Core Mathematics 2)
Year	2016
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Trigonometric equations in context
Type	Solve shifted trig equation
Difficulty	Moderate -0.3 Part (i) is a straightforward rearrangement to cos(θ - π/5) = 1/2 followed by standard angle solving with a phase shift. Part (ii) requires the identity cos²x = 1 - sin²x to form a quadratic in sin x, then solving—a standard C2 technique. Both parts are routine applications of core methods with no novel problem-solving required, making this slightly easier than average.
Spec	1.05o Trigonometric equations: solve in given intervals

6. (i) Solve, for $- \pi < \theta \leqslant \pi$, $$1 - 2 \cos \left( \theta - \frac { \pi } { 5 } \right) = 0$$ giving your answers in terms of $\pi$.
(ii) Solve, for $0 \leqslant x < 360 ^ { \circ }$, $$4 \cos ^ { 2 } x + 7 \sin x - 2 = 0$$ giving your answers to one decimal place.
(Solutions based entirely on graphical or numerical methods are not acceptable.)

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$1 - 2\cos\left(\theta - \frac{\pi}{5}\right) = 0$; $-\pi < \theta \leq \pi$
(i) $\cos\left(\theta - \frac{\pi}{5}\right) = \frac{1}{2}$	M1	Rearranges to give $\cos\left(\theta - \frac{\pi}{5}\right) = \pm\frac{1}{2}$ or $\frac{1}{2}$
$\theta - \frac{\pi}{5} = -\frac{2\pi}{15}$ or $\frac{8\pi}{15}$ or $-24°$ or $96°$ or awrt 1.68 or awrt -0.419	A1	At least one of $-\frac{2\pi}{15}$ or $\frac{8\pi}{15}$ or $-24°$ or $96°$ or awrt 1.68 or awrt -0.419
$\theta = \left\{-\frac{2\pi}{15}, \frac{8\pi}{15}\right\}$	A1	Both $-\frac{2\pi}{15}$ and $\frac{8\pi}{15}$
[3]
NB Misread		Misreading $\frac{\pi}{2}$ as $\frac{\pi}{6}$ or $\frac{\pi}{3}$ (or anything else) – treat as misread so M1 A0 A0 is maximum mark
Across 0 to 180: $\theta + 2 = 0$, $0 \leq \theta < 360°$
(ii) $4(1 - \sin x)^2 + 7\sin x - 2 = 0$	M1	Applies $\cos^2 x = 1 - \sin^2 x$
$4 - 4\sin^2 x + 7\sin x - 2 = 0$
$4\sin^2 x - 7\sin x - 2 = 0$	A1 oe	Correct 3 term equation: $4\sin^2 x - 7\sin x - 2 = \{0\}$
$(4\sin x + 1)(\sin x - 2) = \{0\}$, $\sin x = ...$	M1	Valid attempt at solving and $\sin x = ...$
$\sin x = -\frac{1}{4}$, $\{\sin x = 2\}$	A1 cso	$\sin x = -\frac{1}{4}$ (See notes.)
$x = \text{awrt}[194.5, 345.5]$ or awrt 3.4 or awrt 6.0	A1ft	At least one of awrt 194.5 or awrt 345.5 or awrt 3.4 or awrt 6.0. This is a limited follow-through.
awrt 194.5 and awrt 345.5
A1
[5]

Question 6 Notes:

Answer	Marks
Note	Details
M1	Rearranges to give $\cos\left(\theta - \frac{\pi}{5}\right) = \pm\frac{1}{2}$ or $\frac{1}{2}$
Note	M1 can be implied by seeing either $\frac{\pi}{3}$ or $60°$ as a result of taking $\cos^{-1}(...)$
A1	Answers may be in degrees or radians for this mark and may have just one correct answer before units in working if correct answers follow (recovery)
A1	Both answers correct and in radians as multiples of $\pi$: $-\frac{2\pi}{15}$ and $\frac{8\pi}{15}$. Ignore EXTRA solutions outside the range $-\pi < \theta \leq \pi$ but lose this mark for extra solutions in this range.
NB Misread	Writing equation as $4\cos^2 x - 7\sin x - 2 = 0$ with a sign error should be marked by applying the scheme as it simplifies the solution. (do not treat as misread) Max mark is 3/6
M1	Using $\cos^2 x = 1 - \sin^2 x$ on the given equation. [Applying $\cos^2 x = \sin^2 x - 1$, scores M0.]
1st A1	Obtaining a correct three term equation e.g. either $4\sin^2 x - 7\sin x - 2 = \{0\}$ or $-4\sin^2 x + 7\sin x + 2 = \{0\}$ or $4\sin^2 x - 7\sin x = 2$ or $4\sin^2 x = 7\sin x + 2$ etc.
2nd M1	For a valid attempt at solving a 3TQ quadratic in sine. Methods include factorization, quadratic formula, completion of the square (unlikely here) and calculator. (See notes on page 6 for general principles on awarding this mark) Can use any variable here, $s, y, x$ or $\sin x$, and an attempt to find at least one of the solutions for sinx. This solution may be outside the range for sinx
2nd A1	$\sin x = -\frac{1}{4}$ BY A CORRECT SOLUTION ONLY UP TO THIS POINT. Ignore extra answer of $\sin x = 2$, but penalise if candidate states an incorrect result, e.g. $\sin x = -2$. $\sin x = -\frac{1}{4}$ can be implied by later correct working if no errors are seen.
3rd A1ft	At least one of awrt 194.5 or awrt 345.5 or awrt 3.4 or awrt 6.0. This is a limited follow-through. Only follow through on the error $\sin x = -\frac{1}{4}$ and allow for 165.5 special case (as this is equivalent work). This error is likely to earn M1A1M1A0A0 so 4/6 or M1A0M1A0A0 if the quadratic had a sign slip.
4th A1	awrt 194.5 and awrt 345.5
Note	If there are any EXTRA solutions inside the range $0°, x < 360°$ and the candidate would otherwise score FULL MARKS then withhold the final A1 mark.
Special Cases	Rounding error: Allow M1A1M1A1A0 for those who give two correct answers but wrong accuracy e.g. awrt 194, 346 (Remove final A1 for this error) Answers in radians:– lose final mark so either or both of 3.4, 6.0 gets A1ftA0 It is possible to earn M1A0A1A1 on the final 4 marks if an error results fortuitously in $\sin x = -1/4$ then correct work follows.

| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 - 2\cos\left(\theta - \frac{\pi}{5}\right) = 0$; $-\pi < \theta \leq \pi$ | | |
| **(i)** $\cos\left(\theta - \frac{\pi}{5}\right) = \frac{1}{2}$ | M1 | Rearranges to give $\cos\left(\theta - \frac{\pi}{5}\right) = \pm\frac{1}{2}$ or $\frac{1}{2}$ |
| $\theta - \frac{\pi}{5} = -\frac{2\pi}{15}$ or $\frac{8\pi}{15}$ or $-24°$ or $96°$ or awrt 1.68 or awrt -0.419 | A1 | At least one of $-\frac{2\pi}{15}$ or $\frac{8\pi}{15}$ or $-24°$ or $96°$ or awrt 1.68 or awrt -0.419 |
| $\theta = \left\{-\frac{2\pi}{15}, \frac{8\pi}{15}\right\}$ | A1 | Both $-\frac{2\pi}{15}$ and $\frac{8\pi}{15}$ |
| | [3] | |
| **NB Misread** | | Misreading $\frac{\pi}{2}$ as $\frac{\pi}{6}$ or $\frac{\pi}{3}$ (or anything else) – treat as misread so M1 A0 A0 is maximum mark |
| | | Across 0 to 180: $\theta + 2 = 0$, $0 \leq \theta < 360°$ |
| **(ii)** $4(1 - \sin x)^2 + 7\sin x - 2 = 0$ | M1 | Applies $\cos^2 x = 1 - \sin^2 x$ |
| $4 - 4\sin^2 x + 7\sin x - 2 = 0$ | | |
| $4\sin^2 x - 7\sin x - 2 = 0$ | A1 oe | Correct 3 term equation: $4\sin^2 x - 7\sin x - 2 = \{0\}$ |
| $(4\sin x + 1)(\sin x - 2) = \{0\}$, $\sin x = ...$ | M1 | Valid attempt at solving and $\sin x = ...$ |
| $\sin x = -\frac{1}{4}$, $\{\sin x = 2\}$ | A1 cso | $\sin x = -\frac{1}{4}$ (See notes.) |
| $x = \text{awrt}[194.5, 345.5]$ or awrt 3.4 or awrt 6.0 | A1ft | At least one of awrt 194.5 or awrt 345.5 or awrt 3.4 or awrt 6.0. This is a limited follow-through. |
| | | awrt 194.5 and awrt 345.5 |
| | | A1 |
| | [5] | |

**Question 6 Notes:**

| Note | Details |
|---|---|
| M1 | Rearranges to give $\cos\left(\theta - \frac{\pi}{5}\right) = \pm\frac{1}{2}$ or $\frac{1}{2}$ |
| Note | M1 can be implied by seeing either $\frac{\pi}{3}$ or $60°$ as a result of taking $\cos^{-1}(...)$ |
| A1 | Answers may be in degrees or radians for this mark and may have just one correct answer before units in working if correct answers follow (recovery) |
| A1 | Both answers correct and in radians as multiples of $\pi$: $-\frac{2\pi}{15}$ and $\frac{8\pi}{15}$. Ignore EXTRA solutions outside the range $-\pi < \theta \leq \pi$ but lose this mark for extra solutions in this range. |
| NB Misread | Writing equation as $4\cos^2 x - 7\sin x - 2 = 0$ with a sign error should be marked by applying the scheme as it simplifies the solution. (do not treat as misread) Max mark is 3/6 |
| M1 | Using $\cos^2 x = 1 - \sin^2 x$ on the given equation. [Applying $\cos^2 x = \sin^2 x - 1$, scores M0.] |
| 1st A1 | Obtaining a correct three term equation e.g. either $4\sin^2 x - 7\sin x - 2 = \{0\}$ or $-4\sin^2 x + 7\sin x + 2 = \{0\}$ or $4\sin^2 x - 7\sin x = 2$ or $4\sin^2 x = 7\sin x + 2$ etc. |
| 2nd M1 | For a valid attempt at solving a 3TQ quadratic in sine. Methods include factorization, quadratic formula, completion of the square (unlikely here) and calculator. (See notes on page 6 for general principles on awarding this mark) Can use any variable here, $s, y, x$ or $\sin x$, and an attempt to find at least one of the solutions for sinx. This solution may be outside the range for sinx |
| 2nd A1 | $\sin x = -\frac{1}{4}$ BY A CORRECT SOLUTION ONLY UP TO THIS POINT. Ignore extra answer of $\sin x = 2$, but penalise if candidate states an incorrect result, e.g. $\sin x = -2$. $\sin x = -\frac{1}{4}$ can be implied by later correct working if no errors are seen. |
| 3rd A1ft | At least one of awrt 194.5 or awrt 345.5 or awrt 3.4 or awrt 6.0. This is a limited follow-through. Only follow through on the error $\sin x = -\frac{1}{4}$ and allow for 165.5 special case (as this is equivalent work). This error is likely to earn M1A1M1A0A0 so 4/6 or M1A0M1A0A0 if the quadratic had a sign slip. |
| 4th A1 | awrt 194.5 and awrt 345.5 |
| Note | If there are any EXTRA solutions inside the range $0°, x < 360°$ and the candidate would otherwise score FULL MARKS then withhold the final A1 mark. |
| Special Cases | Rounding error: Allow M1A1M1A1A0 for those who give two correct answers but wrong accuracy e.g. awrt 194, 346 (Remove final A1 for this error) Answers in radians:– lose final mark so either or both of 3.4, 6.0 gets A1ftA0 It is possible to earn M1A0A1A1 on the final 4 marks if an error results fortuitously in $\sin x = -1/4$ then correct work follows. |

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Show LaTeX source

6. (i) Solve, for $- \pi < \theta \leqslant \pi$,

$$1 - 2 \cos \left( \theta - \frac { \pi } { 5 } \right) = 0$$

giving your answers in terms of $\pi$.\\
(ii) Solve, for $0 \leqslant x < 360 ^ { \circ }$,

$$4 \cos ^ { 2 } x + 7 \sin x - 2 = 0$$

giving your answers to one decimal place.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)

\hfill \mbox{\textit{Edexcel C2 2016 Q6 [9]}}

This paper (9 questions)

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Q1 7 Q2 6 Q3 8 Q4 8 Q5 9 Q6 9 Q7 6 Q8 7 Q9 15

Answer	Marks
Note	Details
M1	Rearranges to give \(\cos\left(\theta - \frac{\pi}{5}\right) = \pm\frac{1}{2}\) or \(\frac{1}{2}\)
Note	M1 can be implied by seeing either \(\frac{\pi}{3}\) or \(60°\) as a result of taking \(\cos^{-1}(...)\)
A1	Answers may be in degrees or radians for this mark and may have just one correct answer before units in working if correct answers follow (recovery)
A1	Both answers correct and in radians as multiples of \(\pi\): \(-\frac{2\pi}{15}\) and \(\frac{8\pi}{15}\). Ignore EXTRA solutions outside the range \(-\pi < \theta \leq \pi\) but lose this mark for extra solutions in this range.
NB Misread	Writing equation as \(4\cos^2 x - 7\sin x - 2 = 0\) with a sign error should be marked by applying the scheme as it simplifies the solution. (do not treat as misread) Max mark is 3/6
M1	Using \(\cos^2 x = 1 - \sin^2 x\) on the given equation. [Applying \(\cos^2 x = \sin^2 x - 1\), scores M0.]
1st A1	Obtaining a correct three term equation e.g. either \(4\sin^2 x - 7\sin x - 2 = \{0\}\) or \(-4\sin^2 x + 7\sin x + 2 = \{0\}\) or \(4\sin^2 x - 7\sin x = 2\) or \(4\sin^2 x = 7\sin x + 2\) etc.
2nd M1	For a valid attempt at solving a 3TQ quadratic in sine. Methods include factorization, quadratic formula, completion of the square (unlikely here) and calculator. (See notes on page 6 for general principles on awarding this mark) Can use any variable here, \(s, y, x\) or \(\sin x\), and an attempt to find at least one of the solutions for sinx. This solution may be outside the range for sinx
2nd A1	\(\sin x = -\frac{1}{4}\) BY A CORRECT SOLUTION ONLY UP TO THIS POINT. Ignore extra answer of \(\sin x = 2\), but penalise if candidate states an incorrect result, e.g. \(\sin x = -2\). \(\sin x = -\frac{1}{4}\) can be implied by later correct working if no errors are seen.
3rd A1ft	At least one of awrt 194.5 or awrt 345.5 or awrt 3.4 or awrt 6.0. This is a limited follow-through. Only follow through on the error \(\sin x = -\frac{1}{4}\) and allow for 165.5 special case (as this is equivalent work). This error is likely to earn M1A1M1A0A0 so 4/6 or M1A0M1A0A0 if the quadratic had a sign slip.
4th A1	awrt 194.5 and awrt 345.5
Note	If there are any EXTRA solutions inside the range \(0°, x < 360°\) and the candidate would otherwise score FULL MARKS then withhold the final A1 mark.
Special Cases	Rounding error: Allow M1A1M1A1A0 for those who give two correct answers but wrong accuracy e.g. awrt 194, 346 (Remove final A1 for this error) Answers in radians:– lose final mark so either or both of 3.4, 6.0 gets A1ftA0 It is possible to earn M1A0A1A1 on the final 4 marks if an error results fortuitously in \(\sin x = -1/4\) then correct work follows.