C2 January 2007 Q6
6. Find all the solutions, in the interval \(0 \leqslant x \lt 2\pi\), of the equation\[2\cos^2 x + 1 = 5\sin x,\]giving each solution in terms of \(\pi\). (6)
| Scheme | Marks |
|---|---|
| \(2(1 - \sin^2 x) + 1 = 5\sin x\) | M1 |
| \(2\sin^2 x + 5\sin x - 3 = 0\) \((2\sin x - 1)(\sin x + 3) = 0\) | |
| \(\sin x = \dfrac{1}{2}\) | M1, A1 |
| \(x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}\) | M1, M1, A1cso |
| (6) |
Notes
M1: Use of \(\cos^2 x = 1 - \sin^2 x\).
Condone invisible brackets in first line if \(2 - 2\sin^2 x\) is present (or implied) in a subsequent line.
Must be using \(\cos^2 x = 1 - \sin^2 x\). Using \(\cos^2 x = 1 + \sin^2 x\) is M0.
M1: Attempt to solve a 2 or 3 term quadratic in \(\sin x\) up to \(\sin x = \ldots\)
Usual rules for solving quadratics. Method may be factorising, formula or completing the square
A1: Correct factorising for correct quadratic and \(\sin x = \dfrac{1}{2}\).
So, e.g. \((\sin x + 3)\) as a factor \(\to \sin x = 3\) can be ignored.
M1: Method for finding any angle in any range consistent with (either of) their trig. equation(s) in degrees or radians (even if \(x\) not exact). [Generous M mark]
Generous mark. Solving any trig. equation that comes from minimal working (however bad).
So \(x = \sin^{-1}/\cos^{-1}/\tan^{-1}(\text{number}) \to\) answer in degrees or radians correct for their equation (in any range)
M1: Method for finding second angle consistent with (either of) their trig. equation(s) in radians.
Must be in range \(0 \leqslant x \lt 2\pi\). Must involve using \(\pi\) (e.g. \(\pi \pm \ldots, 2\pi - \ldots\)) but … can be inexact.
Must be using the same equation as they used to attempt the 3rd M mark.
Use of \(\pi\) must be consistent with the trig. equation they are using (e.g. if using \(\cos^{-1}\) then must be using \(2\pi - \ldots\))
If finding both angles in degrees: method for finding 2nd angle equivalent to method above in degrees and an attempt to change both angles to radians.
A1 cso: \(\dfrac{\pi}{6}, \dfrac{5\pi}{6}\) c.s.o. Recurring decimals are okay (instead of \(\dfrac{1}{6}\) and \(\dfrac{5}{6}\)).
Correct decimal values (corrected or truncated) before the final answer of \(\dfrac{\pi}{6}, \dfrac{5\pi}{6}\) is acceptable.
Ignore extra solutions outside range; deduct final A mark for extra solutions in range.
Special case
Answer only \(\dfrac{\pi}{6}, \dfrac{5\pi}{6}\) M0, M0, A0, M1, M1 A1
Answer only \(\dfrac{\pi}{6}\) M0, M0, A0, M1, M0 A0
Finding answers by trying different values (e.g. trying multiples of \(\pi\)) in \(2\cos^2 x + 1 = 5\sin x\): as for answer only.