C3 January 2011 Q3
3. Find all the solutions of
\[2\cos 2\theta = 1 - 2\sin\theta\]in the interval \(0 \leqslant \theta \lt 360^\circ\). (6)
| Scheme | Marks |
|---|---|
| \(2\cos 2\theta = 1 - 2\sin\theta\) | |
| \(2(1 - 2\sin^2\theta) = 1 - 2\sin\theta\) | M1 |
| \(2 - 4\sin^2\theta = 1 - 2\sin\theta\) | |
| \(4\sin^2\theta - 2\sin\theta - 1 = 0\) | M1(*) |
| \(\sin\theta = \dfrac{2 \pm \sqrt{4 - 4(4)(-1)}}{8}\) | M1 |
| PVs: \(\alpha_1 = 54^\circ\) or \(\alpha_2 = -18^\circ\) | |
| \(\theta = \{54, 126, 198, 342\}\) | A1 dM1(*) A1 |
| (6 marks) |
Notes
M1: Substitutes either \(1 - 2\sin^2\theta\) or \(2\cos^2\theta - 1\) or \(\cos^2\theta - \sin^2\theta\) for \(\cos 2\theta\).
M1(*): Forms a “quadratic in sine” \(= 0\)
M1: Applies the quadratic formula. See notes for alternative methods.
A1: Any one correct answer
dM1(*): 180-their pv
A1: All four solutions correct.