Standard +0.3 This is a structured two-part question where the first part (showing the rearrangement) guides students through multiplying by sin x and using the Pythagorean identity, making it more accessible than if they had to discover the approach themselves. The second part is routine quadratic solving followed by standard inverse trig. Slightly above average due to the reciprocal trig manipulation, but the scaffolding and standard techniques keep it accessible.
2 Show that the equation \(\operatorname { cosec } x + 5 \cot x = 3 \sin x\) may be rearranged as
$$3 \cos ^ { 2 } x + 5 \cos x - 2 = 0$$
Hence solve the equation for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\), giving your answers to 1 decimal place.
Using \(\cosec x = 1/\sin x\) and \(\cot x = \cos x / \sin x\)
\(\Rightarrow 1 + 5\cos x = 3\sin^2 x = 3(1 - \cos^2 x)\)
M1
\(\cos^2 x + \sin^2 x = 1\) used (both M marks must be part of same solution to score both)
\(\Rightarrow 3\cos^2 x + 5\cos x - 2 = 0\) *
A1
AG (Accept working backwards, with same stages needed)
\(\Rightarrow (3\cos x - 1)(\cos x + 2) = 0\)
M1
Use of correct quadratic formula (can be error when substituting into correct formula) or factorising (giving correct coeffs 3 and -2 when multiplied out) or comp square oe
\(\Rightarrow \cos x = 1/3\)
A1
\(\cos x = 1/3\) www
\(x = 70.5°\)
A1
For \(70.5°\) or first correct solution, condone over-specification (ie \(70.5°\) or better eg \(70.53°, 70.5288°\) etc)
\(289.5°\)
A1
For \(289.5°\) or second correct solution (condone over-specification) and no others in the range. Ignore solutions outside the range. SCA1A0 for incorrect answers that round to 70.5 and 360-their ans, eg 70.52 and 289.48. SC Award A1A0 for 1.2, 5.1 radians (or better). Do not award SC marks if there are extra solutions in the range.
[7]
## Question 2:
$\cosec x + 5\cot x = 3\sin x$
| Answer | Mark | Guidance |
|--------|------|----------|
| $\Rightarrow \frac{1}{\sin x} + \frac{5\cos x}{\sin x} = 3\sin x$ | M1 | Using $\cosec x = 1/\sin x$ and $\cot x = \cos x / \sin x$ |
| $\Rightarrow 1 + 5\cos x = 3\sin^2 x = 3(1 - \cos^2 x)$ | M1 | $\cos^2 x + \sin^2 x = 1$ **used** (both M marks must be part of same solution to score both) |
| $\Rightarrow 3\cos^2 x + 5\cos x - 2 = 0$ * | A1 | **AG** (**Accept working backwards, with same stages needed**) |
| $\Rightarrow (3\cos x - 1)(\cos x + 2) = 0$ | M1 | Use of correct quadratic formula (can be error when substituting into correct formula) or factorising (giving correct coeffs 3 and -2 when multiplied out) or comp square oe |
| $\Rightarrow \cos x = 1/3$ | A1 | $\cos x = 1/3$ www |
| $x = 70.5°$ | A1 | For $70.5°$ or first correct solution, condone over-specification (ie $70.5°$ or better eg $70.53°, 70.5288°$ etc) |
| $289.5°$ | A1 | For $289.5°$ or second correct solution (condone over-specification) and no others in the range. Ignore solutions outside the range. **SC**A1A0 for incorrect answers that round to 70.5 and 360-their ans, eg 70.52 and 289.48. **SC** Award A1A0 for 1.2, 5.1 radians (or better). Do not award SC marks if there are extra solutions in the range. |
| | **[7]** | |
---
2 Show that the equation $\operatorname { cosec } x + 5 \cot x = 3 \sin x$ may be rearranged as
$$3 \cos ^ { 2 } x + 5 \cos x - 2 = 0$$
Hence solve the equation for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$, giving your answers to 1 decimal place.
\hfill \mbox{\textit{OCR MEI C4 Q2 [7]}}