C3 January 2012 Q5
5. Solve, for \(0 \leqslant \theta \leqslant 180^\circ\),
\[2\cot^2 3\theta = 7\operatorname{cosec} 3\theta - 5\]
Give your answers in degrees to 1 decimal place. (10)
| Scheme | Marks |
|---|---|
| Uses the identity \(\cot^2(3\theta) = \operatorname{cosec}^2(3\theta) - 1\) in \(2\cot^2(3\theta) = 7\operatorname{cosec}(3\theta) - 5\) | M1 |
| \(2\operatorname{cosec}^2(3\theta) - 7\operatorname{cosec}(3\theta) + 3 = 0\) | A1 |
| \((2\operatorname{cosec}3\theta - 1)(\operatorname{cosec}3\theta - 3) = 0\) | dM1 |
| \(\operatorname{cosec}3\theta = 3\) | A1 |
| \(\theta = \dfrac{\text{invsin}\left(\frac{1}{3}\right)}{3},\ \dfrac{19.5^\circ}{3} = \text{awrt } 6.5^\circ\) | ddM1, A1 |
| \(\theta = \dfrac{180^\circ - \text{invsin}\left(\frac{1}{3}\right)}{3}, 53.5^\circ\) Correct 2nd value | ddM1,A1 |
| \(\theta = \dfrac{360^\circ + \text{invsin}\left(\frac{1}{3}\right)}{3}\) Correct 3rd value | ddM1 |
| All 4 correct answers awrt \(6.5^\circ, 53.5^\circ, 126.5^\circ\) or \(173.5^\circ\) | A1 |
| (10 marks) |
Notes
M1 Uses the substitution \(\cot^2(3\theta) = \pm 1 \pm \operatorname{cosec}^2(3\theta)\) to produce a quadratic equation in \(\operatorname{cosec}(3\theta)\)
Accept ‘invisible’ brackets in which \(2\cot^2(3\theta)\) is replaced by \(2\operatorname{cosec}^2(3\theta) - 1\)
A (longer) but acceptable alternative is to convert everything to \(\sin(3\theta)\).
For this to be scored \(\cot^2 3\theta\) must be replaced by \(\dfrac{\cos^2(3\theta)}{\sin^2(3\theta)}\), \(\operatorname{cosec}(3\theta)\) must be replaced by \(\dfrac{1}{\sin 3\theta}\).
An attempt must be made to multiply by \(\sin^2(3\theta)\) and finally \(\cos^2(3\theta)\) replaced by \(= \pm 1 \pm \sin^2(3\theta)\)
A1 A correct equation (=0) written or implied by working is obtained. Terms must be collected together on one side of the equation. The usual alternatives are
\(2\operatorname{cosec}^2(3\theta) - 7\operatorname{cosec}(3\theta) + 3 = 0\) or \(3\sin^2(3\theta) - 7\sin(3\theta) + 2 = 0\)
dM1 Either an attempt to factorise a 3 term quadratic in \(\boldsymbol{\operatorname{cosec}(3\theta)}\) or \(\boldsymbol{\sin(3\theta)}\) with the usual rules
Or use of a correct formula to produce a solution in \(\operatorname{cosec}(3\theta)\) or \(\sin(3\theta)\)
A1 Obtaining the correct value of \(\operatorname{cosec}(3\theta) = 3\) or \(\sin(3\theta) = \frac{1}{3}\). Ignore other values
ddM1 Correct method to produce the principal value of \(\theta\). It is dependent upon the two M’s being scored.
Look for \(\theta = \dfrac{\text{invsin}\left(\text{their } \frac{1}{3}\right)}{3}\)
A1 Awrt 6.5
ddM1 Correct method to produce a secondary value. This is dependent upon the candidate having scored the first 2 M’s. Usually you look for \(\dfrac{180 - \text{their } 19.5}{3}\) or \(\dfrac{360 + \text{their } 19.5}{3}\) or \(\dfrac{540 - \text{their } 19.5}{3}\)
Note 180-their 6.5 must be marked correct BUT 360+their 6.5 is incorrect
A1 Any other correct answer. Awrt 6.5,53.5,126.5 or 173.5
ddM1 Correct method to produce a THIRD value. This is dependent upon the candidate having scored the first 2 M’s. See above for alternatives
A1 All 4 correct answers awrt 6.5,53.5,126.5 or 173.5 and no extras inside the range. Ignore any answers outside the range.
Radian answers: awrt 0.11, 0.93, 2.21, 3.03. Accuracy must be to 2dp.
Lose the first mark that could have been scored. Fully correct radian answer scores 1,1,1,1,1,0,1,1,1,1=9 marks
Candidates cannot mix degrees and radians for method marks.
Special case: Some candidates solve the equation in \(\boldsymbol{\operatorname{cosec}(\theta \text{ or } x)}\), \(\boldsymbol{\sin(\theta \text{ or } x)}\) to produce \(\boldsymbol{\operatorname{cosec}(\theta \text{ or } x) = 3}\) \(\boldsymbol{\sin(\theta \text{ or } x) = \frac{1}{3}}\).