| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Solve shifted trig equation |
| Difficulty | Standard +0.3 Part (i) is a straightforward linear sine equation requiring standard techniques (isolate, inverse function, adjust for multiple solutions within range). Part (ii) requires converting tan to sin/cos, multiplying through to clear denominators, and using the Pythagorean identity to form a quadratic in cos θ—this is a standard P2 technique but involves more algebraic manipulation. Both parts are typical textbook exercises with clear solution paths, slightly above average due to the multi-step algebraic work in part (ii). |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(5\sin(3x+0.1)+2=0 \Rightarrow \sin(3x+0.1)=-\frac{2}{5}\) | M1 | For \(\pm2\) then dividing by 5 to reach \(\sin(3x+0.1)=\pm\frac{2}{5}\). May be implied by \(3x+0.1=\pm0.411...\). Condone working in degrees for this mark only. |
| \(3x+0.1=\sin^{-1}\left(-\frac{2}{5}\right) \Rightarrow x=\frac{\sin^{-1}\left(-\frac{2}{5}\right)-0.1}{3}\) | dM1 | Correct strategy for finding \(x\). Allow \(x=\frac{\pm2\pi n+\sin^{-1}\left(\pm\frac{2}{5}\right)\pm0.1}{3}\) or \(x=\frac{\pi-\sin^{-1}\left(\pm\frac{2}{5}\right)\pm0.1}{3}\). Must reach \(\sin(3x+0.1)=-\frac{2}{5}\) before achieving an angle. Must be working in radians OR entirely in degrees. Dependent on first M1. |
| \(x=-0.94,\ -0.17,\ 1.15,\ 1.92\) | A1 | Two of awrt \(-0.94,\ -0.17,\ 1.15,\ 1.92\). Must be in radians. |
| A1 | All of awrt \(-0.94,\ -0.17,\ 1.15,\ 1.92\) and no extras in range. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(2\tan\theta\sin\theta=\cos\theta+5 \Rightarrow 2\sin^2\theta=\cos^2\theta+5\cos\theta\) | M1 | For using \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) and multiplying by \(\cos\theta\). Look for denominator being removed from \(\frac{\sin^2\theta}{\cos\theta}\) and multiplying at least one other term by \(\cos\theta\). |
| \(\Rightarrow 2(1-\cos^2\theta)=\cos^2\theta+5\cos\theta\) | M1 | Attempts to use \(\pm\sin^2\theta=\pm1\pm\cos^2\theta\) and proceeds to equation in \(\cos\theta\) only. |
| \(\Rightarrow 3\cos^2\theta+5\cos\theta-2=0\) | A1 | Not all terms need to be on same side. Condone omission of \(=0\) if all on one side. Condone poor notation e.g. \(\cos\theta^2\). |
| \(\cos\theta=\frac{1}{3},\ (-2) \Rightarrow \theta=\cos^{-1}\left(\frac{1}{3}\right)=\ldots\) | M1 | Solves their 3TQ in \(\cos\theta\) and takes inverse cos of one root to obtain at least one value. Condone angles in radians (awrt 1.23 or awrt 5.05). |
| \((\theta=)\ 70.5°,\ 289.5°\) | A1 | awrt 70.5, awrt 289.5 and no others in range. Must be in degrees not radians. |
## Question 8(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $5\sin(3x+0.1)+2=0 \Rightarrow \sin(3x+0.1)=-\frac{2}{5}$ | M1 | For $\pm2$ then dividing by 5 to reach $\sin(3x+0.1)=\pm\frac{2}{5}$. May be implied by $3x+0.1=\pm0.411...$. Condone working in degrees for this mark only. |
| $3x+0.1=\sin^{-1}\left(-\frac{2}{5}\right) \Rightarrow x=\frac{\sin^{-1}\left(-\frac{2}{5}\right)-0.1}{3}$ | dM1 | Correct strategy for finding $x$. Allow $x=\frac{\pm2\pi n+\sin^{-1}\left(\pm\frac{2}{5}\right)\pm0.1}{3}$ or $x=\frac{\pi-\sin^{-1}\left(\pm\frac{2}{5}\right)\pm0.1}{3}$. Must reach $\sin(3x+0.1)=-\frac{2}{5}$ before achieving an angle. Must be working in radians OR entirely in degrees. Dependent on first M1. |
| $x=-0.94,\ -0.17,\ 1.15,\ 1.92$ | A1 | Two of awrt $-0.94,\ -0.17,\ 1.15,\ 1.92$. Must be in radians. |
| | A1 | All of awrt $-0.94,\ -0.17,\ 1.15,\ 1.92$ and no extras in range. |
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## Question 8(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $2\tan\theta\sin\theta=\cos\theta+5 \Rightarrow 2\sin^2\theta=\cos^2\theta+5\cos\theta$ | M1 | For using $\tan\theta=\frac{\sin\theta}{\cos\theta}$ and multiplying by $\cos\theta$. Look for denominator being removed from $\frac{\sin^2\theta}{\cos\theta}$ and multiplying at least one other term by $\cos\theta$. |
| $\Rightarrow 2(1-\cos^2\theta)=\cos^2\theta+5\cos\theta$ | M1 | Attempts to use $\pm\sin^2\theta=\pm1\pm\cos^2\theta$ and proceeds to equation in $\cos\theta$ only. |
| $\Rightarrow 3\cos^2\theta+5\cos\theta-2=0$ | A1 | Not all terms need to be on same side. Condone omission of $=0$ if all on one side. Condone poor notation e.g. $\cos\theta^2$. |
| $\cos\theta=\frac{1}{3},\ (-2) \Rightarrow \theta=\cos^{-1}\left(\frac{1}{3}\right)=\ldots$ | M1 | Solves their 3TQ in $\cos\theta$ and takes inverse cos of one root to obtain at least one value. Condone angles in radians (awrt 1.23 or awrt 5.05). |
| $(\theta=)\ 70.5°,\ 289.5°$ | A1 | awrt 70.5, awrt 289.5 and no others in range. Must be in degrees not radians. |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions based entirely on calculator technology are not acceptable.\\
(i) Solve, for $- \frac { \pi } { 2 } < x < \pi$, the equation
$$5 \sin ( 3 x + 0.1 ) + 2 = 0$$
giving your answers, in radians, to 2 decimal places.\\
(ii) Solve, for $0 < \theta < 360 ^ { \circ }$, the equation
$$2 \tan \theta \sin \theta = 5 + \cos \theta$$
giving your answers, in degrees, to one decimal place.
\hfill \mbox{\textit{Edexcel P2 2023 Q8 [9]}}