| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: compound angle substitution |
| Difficulty | Moderate -0.3 This is a structured multi-part question with clear scaffolding. Part (a) requires standard algebraic manipulation using tan x = sin x/cos x and the Pythagorean identity—routine for AS level. Part (b) involves solving a quadratic in sin x and finding angles, which is standard practice. Part (c) tests understanding of composite angle transformations but is straightforward once the pattern is recognized. The question is slightly easier than average due to its guided structure and use of well-practiced techniques, though it does require careful algebraic work and understanding of multiple solutions. |
| 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/Working | Mark | Guidance |
| States/uses \(\tan x = \frac{\sin x}{\cos x}\) | B1 | Allow use of \(\theta\); final mark requires equation in terms of \(x\) |
| \(4\sin x = 5\cos^2 x \Rightarrow 4\sin x = 5(1-\sin^2 x)\) | M1 | Multiplies by \(\cos x\); uses \(\cos^2 x = 1 - \sin^2 x\) to form quadratic in \(\sin x\) |
| \(5\sin^2 x + 4\sin x - 5 = 0\) | A1* | Must show all key steps; \(= 0\) must be present in final line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to solve \(5\sin^2 x + 4\sin x - 5 = 0 \Rightarrow \sin x = \ldots\) | M1 | Allow solutions from calculator if one is correct (\(0.6\), \(0.7\), \(-1.4\), or \(-1.5\)) |
| \(\sin x = \frac{-2 \pm \sqrt{29}}{5}\) (\(\sin x =\) awrt \(0.677\)) | A1 | Also accept \(\sin x = \frac{-4 \pm \sqrt{116}}{10}\) |
| Takes \(\sin^{-1}\) leading to at least one answer in range | dM1 | If \(\sin x = 0.677\): expect awrt \(0.744\) or awrt \(2.40\) in radians |
| \(x =\) awrt \(42.6°\) and \(x =\) awrt \(137.4°\) only | A1 | Ignore values outside \(0\) to \(360°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(15 \times 2 = 30\) following through on their "2" | B1ft | Follow through on 15 multiplied by number of solutions in (b) in range \(0\) to \(360°\) |
| Explains mathematically: \(3 \times 5 \times\) their number in range \(0\) to \(360°\); or in words: \(3 \times\) "2" values every \(360°\) and \(5\) lots of \(360°\) | B1ft | Also accept: \(\frac{5400}{360} = 15\) and \(15 \times 2 = 30\) |
# Question 12:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| States/uses $\tan x = \frac{\sin x}{\cos x}$ | B1 | Allow use of $\theta$; final mark requires equation in terms of $x$ |
| $4\sin x = 5\cos^2 x \Rightarrow 4\sin x = 5(1-\sin^2 x)$ | M1 | Multiplies by $\cos x$; uses $\cos^2 x = 1 - \sin^2 x$ to form quadratic in $\sin x$ |
| $5\sin^2 x + 4\sin x - 5 = 0$ | A1* | Must show all key steps; $= 0$ must be present in final line |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to solve $5\sin^2 x + 4\sin x - 5 = 0 \Rightarrow \sin x = \ldots$ | M1 | Allow solutions from calculator if one is correct ($0.6$, $0.7$, $-1.4$, or $-1.5$) |
| $\sin x = \frac{-2 \pm \sqrt{29}}{5}$ ($\sin x =$ awrt $0.677$) | A1 | Also accept $\sin x = \frac{-4 \pm \sqrt{116}}{10}$ |
| Takes $\sin^{-1}$ leading to at least one answer in range | dM1 | If $\sin x = 0.677$: expect awrt $0.744$ or awrt $2.40$ in radians |
| $x =$ awrt $42.6°$ and $x =$ awrt $137.4°$ only | A1 | Ignore values outside $0$ to $360°$ |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $15 \times 2 = 30$ following through on their "2" | B1ft | Follow through on 15 multiplied by number of solutions in (b) in range $0$ to $360°$ |
| Explains mathematically: $3 \times 5 \times$ their number in range $0$ to $360°$; or in words: $3 \times$ "2" values every $360°$ and $5$ lots of $360°$ | B1ft | Also accept: $\frac{5400}{360} = 15$ and $15 \times 2 = 30$ |
---\begin{enumerate}
\item In this question you must show detailed reasoning.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.\\
(a) Show that the equation
$$4 \tan x = 5 \cos x$$
can be written as
$$5 \sin ^ { 2 } x + 4 \sin x - 5 = 0$$
(b) Hence solve, for $0 < x \leqslant 360 ^ { \circ }$
$$4 \tan x = 5 \cos x$$
giving your answers to one decimal place.\\
(c) Hence find the number of solutions of the equation
$$4 \tan 3 x = 5 \cos 3 x$$
in the interval $0 < x \leqslant 1800 ^ { \circ }$, explaining briefly the reason for your answer.
\hfill \mbox{\textit{Edexcel AS Paper 1 2023 Q12 [9]}}