| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Convert sin/cos ratio to tan |
| Difficulty | Moderate -0.3 Part (i) is a straightforward application of dividing by cos and solving tan equation with multiple angle. Part (ii) uses the standard sin²x + cos²x identity to form a quadratic in cos x, then solves for specific values. All techniques are routine C2 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\tan 3\theta = \sqrt{3}\), so \((3\theta) = \frac{\pi}{3}\) | M1 | Obtains \(\frac{\pi}{3}\); may be implied by \(\theta = \frac{\pi}{9}\) in final answer |
| Adds \(\pi\) or \(2\pi\) to previous angle (giving \(\frac{4\pi}{3}\) or \(\frac{7\pi}{3}\)) | M1 | Not dependent on previous mark |
| \(\theta = \frac{\pi}{9}, \frac{4\pi}{9}, \frac{7\pi}{9}\) (all three, no extras in range) | A1 | Three correct answers implies M1M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4(1-\cos^2 x) + \cos x = 4-k\); applies \(\sin^2 x = 1-\cos^2 x\) | M1 | Must be awarded before \(k=3\) is substituted |
| Attempts to solve \(4\cos^2 x - \cos x - k = 0\) | dM1 | Uses formula or completing the square |
| \(\cos x = \frac{1 \pm \sqrt{1+16k}}{8}\) or equivalent | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\cos x = \frac{1 \pm \sqrt{49}}{8} = 1\) and \(-\frac{3}{4}\) | M1 | Substitutes \(k=3\) to obtain two values of \(\cos x\) |
| Obtains two correct values of \(x\) from \(0, 139, 221\) | dM1 | |
| \(x = 0\) and \(139\) and \(221\) (allow awrt 139 and 221); must be in degrees | A1 | All three correct; lose mark for excess answers in range |
# Question 8:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\tan 3\theta = \sqrt{3}$, so $(3\theta) = \frac{\pi}{3}$ | M1 | Obtains $\frac{\pi}{3}$; may be implied by $\theta = \frac{\pi}{9}$ in final answer |
| Adds $\pi$ or $2\pi$ to previous angle (giving $\frac{4\pi}{3}$ or $\frac{7\pi}{3}$) | M1 | Not dependent on previous mark |
| $\theta = \frac{\pi}{9}, \frac{4\pi}{9}, \frac{7\pi}{9}$ (all three, no extras in range) | A1 | Three correct answers implies M1M1A1 |
## Part (ii)(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $4(1-\cos^2 x) + \cos x = 4-k$; applies $\sin^2 x = 1-\cos^2 x$ | M1 | Must be awarded before $k=3$ is substituted |
| Attempts to solve $4\cos^2 x - \cos x - k = 0$ | dM1 | Uses formula or completing the square |
| $\cos x = \frac{1 \pm \sqrt{1+16k}}{8}$ or equivalent | A1 | cao |
## Part (ii)(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\cos x = \frac{1 \pm \sqrt{49}}{8} = 1$ and $-\frac{3}{4}$ | M1 | Substitutes $k=3$ to obtain two values of $\cos x$ |
| Obtains two correct values of $x$ from $0, 139, 221$ | dM1 | |
| $x = 0$ and $139$ and $221$ (allow awrt 139 and 221); must be in degrees | A1 | All three correct; lose mark for excess answers in range |
---8. (i) Solve, for $0 \leqslant \theta < \pi$, the equation
$$\sin 3 \theta - \sqrt { 3 } \cos 3 \theta = 0$$
giving your answers in terms of $\pi$.\\
(ii) Given that
$$4 \sin ^ { 2 } x + \cos x = 4 - k , \quad 0 \leqslant k \leqslant 3$$
\begin{enumerate}[label=(\alph*)]
\item find $\cos x$ in terms of $k$.
\item When $k = 3$, find the values of $x$ in the range $0 \leqslant x < 360 ^ { \circ }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2015 Q8 [9]}}