| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Express and solve equation |
| Difficulty | Standard +0.3 Part (i) requires converting cosec to sin, rearranging to get a standard equation (sin 2θ = √3/8), then solving - straightforward but requires careful algebraic manipulation. Part (ii) is a standard harmonic form question (R cos(x+α) = √3) requiring the textbook method of expressing a cos x + b sin x in R cos(x±α) form, then solving a simple equation. Both parts are routine applications of A-level techniques with no novel insight required, making this slightly easier than average. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cosec\theta = \frac{1}{\sin\theta}\) used to give \(4\cos\theta\sin\theta = \sqrt{3}\) | B1 | For correct use of \(\cosec\theta = \frac{1}{\sin\theta}\) |
| Applies \(\sin 2\theta = 2\sin\theta\cos\theta\) to proceed to \(\sin 2\theta = k\) | M1 | Must reach form \(\sin 2\theta = k\) |
| Uses correct order of operations to find at least one value for \(\theta\) | M1 | Accept radians or degrees |
| \(\theta = 30°, 60°\) | A1 | Both values correct, no extra values in range |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Expresses as \(R\cos(x \pm \alpha) = \sqrt{3}\) | M1 | Correct form attempted |
| \(R = 2\), \(\alpha = \frac{\pi}{3}\), giving \(2\cos\!\left(x + \frac{\pi}{3}\right) = \sqrt{3}\) | M1 | Uses \(R\cos(x+\alpha)\) to find both \(R\) and \(\alpha\) |
| \(\cos\!\left(x + \frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\) | A1 | Correct simplified equation |
| \(x = \arccos\!\left(\frac{\sqrt{3}}{2}\right) - \frac{\pi}{3}\) | M1 | Correct order of operations for at least one value |
| \(x = \frac{3\pi}{2},\ \frac{11\pi}{6}\) | A1 | Both correct, no extra values in range |
## Question 7:
### Part (i): Solve $0 \leq \theta \leq 180°$, the equation $4\cos\theta = \sqrt{3}\cosec\theta$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cosec\theta = \frac{1}{\sin\theta}$ used to give $4\cos\theta\sin\theta = \sqrt{3}$ | B1 | For correct use of $\cosec\theta = \frac{1}{\sin\theta}$ |
| Applies $\sin 2\theta = 2\sin\theta\cos\theta$ to proceed to $\sin 2\theta = k$ | M1 | Must reach form $\sin 2\theta = k$ |
| Uses correct order of operations to find at least one value for $\theta$ | M1 | Accept radians or degrees |
| $\theta = 30°, 60°$ | A1 | Both values correct, no extra values in range |
**(4 marks)**
---
### Part (ii): Solve $0 \leq x \leq 2\pi$, the equation $\cos x - \sqrt{3}\sin x = \sqrt{3}$
| Answer/Working | Mark | Guidance |
|---|---|---|
| Expresses as $R\cos(x \pm \alpha) = \sqrt{3}$ | M1 | Correct form attempted |
| $R = 2$, $\alpha = \frac{\pi}{3}$, giving $2\cos\!\left(x + \frac{\pi}{3}\right) = \sqrt{3}$ | M1 | Uses $R\cos(x+\alpha)$ to find both $R$ and $\alpha$ |
| $\cos\!\left(x + \frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ | A1 | Correct simplified equation |
| $x = \arccos\!\left(\frac{\sqrt{3}}{2}\right) - \frac{\pi}{3}$ | M1 | Correct order of operations for at least one value |
| $x = \frac{3\pi}{2},\ \frac{11\pi}{6}$ | A1 | Both correct, no extra values in range |
**(5 marks)**
---\begin{enumerate}
\item (i) Solve $0 \leq \theta \leq 180 ^ { 0 }$, the equation
\end{enumerate}
$$4 \cos \theta = \sqrt { 3 } \operatorname { cosec } \theta$$
(ii) Solve, for $0 \leq x \leq 2 \pi$, the equation
$$\cos x - \sqrt { 3 } \sin x = \sqrt { 3 }$$
\hfill \mbox{\textit{Edexcel PMT Mocks Q7 [9]}}