C3 January 2010 Q8
8. Solve
\[\operatorname{cosec}^2 2x - \cot 2x = 1\]for \(0 \leqslant x \leqslant 180^\circ\). (7)
| Scheme | Marks |
|---|---|
| \(\operatorname{cosec}^2 2x - \cot 2x = 1\), (eqn \(*\)) \(\quad 0 \leqslant x \leqslant 180^\circ\) | |
| Using \(\operatorname{cosec}^2 2x = 1 + \cot^2 2x\) gives \(1 + \cot^2 2x - \cot 2x = 1\) | M1 |
| \(\underline{\cot^2 2x - \cot 2x = 0}\) or \(\cot^2 2x = \cot 2x\) | A1 |
| \(\cot 2x(\cot 2x - 1) = 0\) or \(\cot 2x = 1\) | dM1 |
| \(\cot 2x = 0\) or \(\cot 2x = 1\) | A1 |
| \(\cot 2x = 0 \Rightarrow (\tan 2x \to \infty) \Rightarrow 2x = 90, 270\) \(\Rightarrow x = 45, 135\) \(\cot 2x = 1 \Rightarrow \tan 2x = 1 \Rightarrow 2x = 45, 225\) \(\Rightarrow x = 22.5, 112.5\) | ddM1 |
| Overall, \(x = \{22.5, 45, 112.5, 135\}\) | A1 B1 |
| (7 marks) |
Notes
M1: Writing down or using \(\operatorname{cosec}^2 2x = \pm 1 \pm \cot^2 2x\) or \(\operatorname{cosec}^2\theta = \pm 1 \pm \cot^2\theta\).
A1: For either \(\underline{\cot^2 2x - \cot 2x}\ \{= 0\}\) or \(\cot^2 2x = \cot 2x\)
dM1: Attempt to factorise or solve a quadratic (See rules for factorising quadratics) or cancelling out \(\cot 2x\) from both sides.
A1: Both \(\cot 2x = 0\) and \(\cot 2x = 1\).
ddM1: Candidate attempts to divide at least one of their principal angles by 2. This will be usually implied by seeing \(x = 22.5\) resulting from \(\cot 2x = 1\).
A1: Both \(x = 22.5\) and \(x = 112.5\)
B1: Both \(x = 45\) and \(x = 135\)
If there are any EXTRA solutions inside the range \(0 \leqslant x \leqslant 180^\circ\) and the candidate would otherwise score FULL MARKS then withhold the final accuracy mark (the sixth mark in this question). Also ignore EXTRA solutions outside the range \(0 \leqslant x \leqslant 180^\circ\).