| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: compound angle substitution |
| Difficulty | Standard +0.3 Part (a) is routine algebraic manipulation using tan θ = sin θ/cos θ and sin²θ + cos²θ = 1 to convert to a quadratic in cos θ. Part (b) applies the same result with a double angle substitution and requires solving the quadratic then finding multiple solutions in the given range. This is a standard P2/C3 trigonometric equation question with clear scaffolding, slightly easier than average due to part (a) doing most of the algebraic work. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) to get equation in sine and cosine only: \(3\sin^2\theta + 9\cos\theta = 11 - 5\cos\theta\) or \(3\cos\theta\left(\frac{\sin\theta}{\cos\theta}\sin\theta + 3\right) = 11 - 5\cos\theta\) | M1 | May be in another variable; condone mix of variables |
| Uses \(\pm\sin^2\theta \pm \cos^2\theta = \pm 1\): \(3(1-\cos^2\theta) + 9\cos\theta = 11 - 5\cos\theta\) | dM1 | Dependent on M1; condone slips in manipulation |
| \(3\cos^2\theta - 14\cos\theta + 8 = 0\) * | A1* | No errors including brackets; withhold for poor notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos 2x = \frac{2}{3}\) | B1 | Ignore any reference to 4 |
| \(x = \frac{\cos^{-1}\!\left(\frac{2}{3}\right)}{2} = \ldots\) | M1 | Correct order of operations for at least one value of \(x\) |
| \(x =\) awrt 24.1, awrt 155.9, awrt 204.1, awrt 335.9 | A1A1 | Two correct values for first A1; all four correct and no others for second A1 |
## Question 9:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $\tan\theta = \frac{\sin\theta}{\cos\theta}$ to get equation in sine and cosine only: $3\sin^2\theta + 9\cos\theta = 11 - 5\cos\theta$ or $3\cos\theta\left(\frac{\sin\theta}{\cos\theta}\sin\theta + 3\right) = 11 - 5\cos\theta$ | M1 | May be in another variable; condone mix of variables |
| Uses $\pm\sin^2\theta \pm \cos^2\theta = \pm 1$: $3(1-\cos^2\theta) + 9\cos\theta = 11 - 5\cos\theta$ | dM1 | Dependent on M1; condone slips in manipulation |
| $3\cos^2\theta - 14\cos\theta + 8 = 0$ * | A1* | No errors including brackets; withhold for poor notation |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos 2x = \frac{2}{3}$ | B1 | Ignore any reference to 4 |
| $x = \frac{\cos^{-1}\!\left(\frac{2}{3}\right)}{2} = \ldots$ | M1 | Correct order of operations for at least one value of $x$ |
| $x =$ awrt 24.1, awrt 155.9, awrt 204.1, awrt 335.9 | A1A1 | Two correct values for first A1; all four correct and no others for second A1 |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.\\
(a) Show that
$$3 \cos \theta ( \tan \theta \sin \theta + 3 ) = 11 - 5 \cos \theta$$
may be written as
$$3 \cos ^ { 2 } \theta - 14 \cos \theta + 8 = 0$$
(b) Hence solve, for $0 < x < 360 ^ { \circ }$
$$3 \cos 2 x ( \tan 2 x \sin 2 x + 3 ) = 11 - 5 \cos 2 x$$
giving your answers to one decimal place.
\hfill \mbox{\textit{Edexcel P2 2023 Q9 [7]}}