June 2022 Paper 3 Q4
4 In this question you must show detailed reasoning.
Determine the exact solutions of the equation \(2\cos^2 x = 3\sin x\) for \(0 \leqslant x \leqslant 2\pi\). [5]
| Scheme | Marks | AO |
|---|---|---|
| DR \(2(1 - \sin^2 x) = 3\sin x\) | M1 | 3.1a |
| \(2\sin^2 x + 3\sin x - 2\ [= 0]\) | M1 | 1.1 |
| \(\sin x = \dfrac{1}{2}\) | A1 | 1.1 |
| \(\dfrac{\pi}{6}\) | A1 | 1.1 |
| \(\dfrac{5\pi}{6}\) | B1 | 2.2a |
| [5] |
Notes
M1: For getting a 3-term quadratic on the same side in a single trig ratio (not dep on M1)
A1: BC, ignore second value if presented
A1: First angle correct and in radians
B1: FT (\(\pi -\) their first angle) OR (180 − their first angle) (dep on first M1)
If further solutions in range B0