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 and inverse trig. Slightly above average due to the reciprocal trig manipulation, but the scaffolding and standard techniques keep it accessible.
3 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 \(\csc x = \frac{1}{\sin x}\) and \(\cot x = \frac{\cos x}{\sin x}\)
\(1 + 5\cos x = 3\sin^2 x = 3(1 - \cos^2 x)\)
M1
\(\cos^2 x + \sin^2 x = 1\) used (both M marks must be part of same solution)
\(3\cos^2 x + 5\cos x - 2 = 0\)
A1
AG (Accept working backwards, with same stages needed)
\((3\cos x - 1)(\cos x + 2) = 0\)
M1
use of correct quadratic formula or factorising or completing the square
\(\cos x = \frac{1}{3}\)
A1
\(\cos x = \frac{1}{3}\) www
\(x = 70.5°\)
A1
for \(70.5°\) or first correct solution, condone over-specification
\(289.5°\)
A1
for \(289.5°\) or second correct solution, and no others in the range
## Question 3:
$\csc x + 5\cot x = 3\sin x$ | | |
$\frac{1}{\sin x} + \frac{5\cos x}{\sin x} = 3\sin x$ | M1 | using $\csc x = \frac{1}{\sin x}$ and $\cot x = \frac{\cos x}{\sin x}$ |
$1 + 5\cos x = 3\sin^2 x = 3(1 - \cos^2 x)$ | M1 | $\cos^2 x + \sin^2 x = 1$ used (both M marks must be part of same solution) |
$3\cos^2 x + 5\cos x - 2 = 0$ | A1 | AG (Accept working backwards, with same stages needed) |
$(3\cos x - 1)(\cos x + 2) = 0$ | M1 | use of correct quadratic formula or factorising or completing the square |
$\cos x = \frac{1}{3}$ | A1 | $\cos x = \frac{1}{3}$ www |
$x = 70.5°$ | A1 | for $70.5°$ or first correct solution, condone over-specification |
$289.5°$ | A1 | for $289.5°$ or second correct solution, and no others in the range |
---
3 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 Q3 [7]}}