8 Show that the equation \(4 \cos ^ { 2 } \theta = 4 - \sin \theta\) may be written in the form
$$4 \sin ^ { 2 } \theta - \sin \theta = 0$$
Hence solve the equation \(4 \cos ^ { 2 } \theta = 4 - \sin \theta\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
Show mark scheme
Show mark scheme source
Question 8:
Answer Marks
Guidance
Answer/Working Mark
Guidance
Use of \(\cos^2\theta = 1 - \sin^2\theta\) M1
At least one correct interim step in obtaining \(4\sin^2\theta - \sin\theta = 0\) M1
NB answer given
\(\theta = 0\) and \(180\) B1
\(14.(47\ldots)\) B1
r.o.t to nearest degree or better
\(165-166\) B1
-1 for extras in range
Copy
## Question 8:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $\cos^2\theta = 1 - \sin^2\theta$ | M1 | |
| At least one correct interim step in obtaining $4\sin^2\theta - \sin\theta = 0$ | M1 | NB answer given |
| $\theta = 0$ and $180$ | B1 | |
| $14.(47\ldots)$ | B1 | r.o.t to nearest degree or better |
| $165-166$ | B1 | -1 for extras in range | 5 |
---
Show LaTeX source
Copy
8 Show that the equation $4 \cos ^ { 2 } \theta = 4 - \sin \theta$ may be written in the form
$$4 \sin ^ { 2 } \theta - \sin \theta = 0$$
Hence solve the equation $4 \cos ^ { 2 } \theta = 4 - \sin \theta$ for $0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$.
\hfill \mbox{\textit{OCR MEI C2 Q8 [5]}}