| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Solve shifted trig equation |
| Difficulty | Moderate -0.3 Part (a) is a straightforward phase-shifted sine equation requiring calculator use and consideration of two solutions in the range. Part (b) involves a standard identity substitution (sin²x = 1 - cos²x) leading to a quadratic in cos x, then finding solutions—routine C2 technique but requires multiple steps. Both are textbook-standard exercises with no novel insight needed, making this slightly easier than average. |
| Spec | 1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\sin(x+45°) = \frac{2}{3}\), so \((x+45°) = 41.8103...\) | M1 | \(\sin^{-1}\left(\frac{2}{3}\right)\) or awrt 41.8 or awrt 0.73° |
| \(x + 45° = \{138.1897...,\ 401.8103...\}\) | M1 | \(x+45°=\) either "180 \(-\alpha\)" or "360\(° +\alpha\)" |
| \(x = \{93.2°,\ 356.8°\}\) | A1 | Either awrt 93.2° or awrt 356.8° |
| Both awrt 93.2° and awrt 356.8° | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2(1-\cos^2 x) + 2 = 7\cos x\) | M1 | Applies \(\sin^2 x = 1 - \cos^2 x\) |
| \(2\cos^2 x + 7\cos x - 4 = 0\) | A1 oe | Correct 3-term equation |
| \((2\cos x - 1)(\cos x + 4) = 0\) | M1 | Valid attempt at solving, \(\cos x = \ldots\) |
| \(\cos x = \frac{1}{2}\), \(\{\cos x = -4\}\) | A1 cso | \(\cos x = \frac{1}{2}\) only |
| \(x = \frac{\pi}{3}\) or 1.04719...° | B1 | Either \(\frac{\pi}{3}\) or awrt 1.05° |
| \(x = \frac{5\pi}{3}\) or 5.23598...° | B1 ft | Either \(\frac{5\pi}{3}\) or awrt 5.24° or \(2\pi - \) their \(\beta\) [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = \text{awrt} -3.2\) | B1 (implied by 1st M1) | Can be implied |
| \(x + 45° = 180° - \alpha\) or \(360° + \alpha\) | M1 | Candidate's \(\alpha\) could be in radians |
| awrt \(93.2°\) or awrt \(356.8°\) | A1 | First two M marks imply M1M1A1 |
| Both awrt \(93.2°\) and awrt \(356.8°\) | A1 | Final mark withheld if extra solutions in range |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(\sin^2 x = 1 - \cos^2 x\) on given equation | M1 | Give bod if bracket omitted when substituting |
| \(2\cos^2 x + 7\cos x - 4 = 0\) or \(-2\cos^2 x - 7\cos x + 4 = 0\) | A1 | Also accept \(2\cos^2 x + 7\cos x = 4\) |
| Valid factorisation/quadratic formula attempt in \(\cos\) | M1 | Can use any variable |
| \(\cos x = \frac{1}{2}\) | A1 | By correct solution only to this point; ignore \(\cos x = -4\) |
| \(x = \frac{\pi}{3}\) or awrt \(1.05^c\) | B1 | |
| \(x = \frac{5\pi}{3}\) or awrt \(5.24^c\) | B1ft | ft from \(2\pi - \beta\) or \(360° - \beta\); withhold if extra solutions in range |
## Question 7:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\sin(x+45°) = \frac{2}{3}$, so $(x+45°) = 41.8103...$ | M1 | $\sin^{-1}\left(\frac{2}{3}\right)$ or awrt 41.8 or awrt 0.73° |
| $x + 45° = \{138.1897...,\ 401.8103...\}$ | M1 | $x+45°=$ either "180 $-\alpha$" or "360$° +\alpha$" |
| $x = \{93.2°,\ 356.8°\}$ | A1 | Either awrt 93.2° or awrt 356.8° |
| Both awrt 93.2° and awrt 356.8° | A1 | **[4]** |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $2(1-\cos^2 x) + 2 = 7\cos x$ | M1 | Applies $\sin^2 x = 1 - \cos^2 x$ |
| $2\cos^2 x + 7\cos x - 4 = 0$ | A1 oe | Correct 3-term equation |
| $(2\cos x - 1)(\cos x + 4) = 0$ | M1 | Valid attempt at solving, $\cos x = \ldots$ |
| $\cos x = \frac{1}{2}$, $\{\cos x = -4\}$ | A1 cso | $\cos x = \frac{1}{2}$ only |
| $x = \frac{\pi}{3}$ or 1.04719...° | B1 | Either $\frac{\pi}{3}$ or awrt 1.05° |
| $x = \frac{5\pi}{3}$ or 5.23598...° | B1 ft | Either $\frac{5\pi}{3}$ or awrt 5.24° or $2\pi - $ their $\beta$ **[6]** |
# Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \text{awrt} -3.2$ | B1 (implied by 1st M1) | Can be implied |
| $x + 45° = 180° - \alpha$ or $360° + \alpha$ | M1 | Candidate's $\alpha$ could be in radians |
| awrt $93.2°$ or awrt $356.8°$ | A1 | First two M marks imply M1M1A1 |
| Both awrt $93.2°$ and awrt $356.8°$ | A1 | Final mark withheld if extra solutions in range |
---
# Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\sin^2 x = 1 - \cos^2 x$ on given equation | M1 | Give bod if bracket omitted when substituting |
| $2\cos^2 x + 7\cos x - 4 = 0$ or $-2\cos^2 x - 7\cos x + 4 = 0$ | A1 | Also accept $2\cos^2 x + 7\cos x = 4$ |
| Valid factorisation/quadratic formula attempt in $\cos$ | M1 | Can use any variable |
| $\cos x = \frac{1}{2}$ | A1 | By correct solution only to this point; ignore $\cos x = -4$ |
| $x = \frac{\pi}{3}$ or awrt $1.05^c$ | B1 | |
| $x = \frac{5\pi}{3}$ or awrt $5.24^c$ | B1ft | ft from $2\pi - \beta$ or $360° - \beta$; withhold if extra solutions in range |
---\begin{enumerate}
\item (a) Solve for $0 \leqslant x < 360 ^ { \circ }$, giving your answers in degrees to 1 decimal place,
\end{enumerate}
$$3 \sin \left( x + 45 ^ { \circ } \right) = 2$$
(b) Find, for $0 \leqslant x < 2 \pi$, all the solutions of
$$2 \sin ^ { 2 } x + 2 = 7 \cos x$$
giving your answers in radians.\\
You must show clearly how you obtained your answers.
\hfill \mbox{\textit{Edexcel C2 2011 Q7 [10]}}