C2 June 2016 Q6

Notes

M1 Rearranges to give \(\cos\left(\theta - \dfrac{\pi}{5}\right) = \pm\dfrac{1}{2}\)

Note M1 can be implied by seeing either \(\dfrac{\pi}{3}\) or \(60^\circ\) as a result of taking \(\cos^{-1}(\ldots)\).

A1 Answers may be in degrees or radians for this mark and may have just one correct answer Ignore mixed units in working if correct answers follow (recovery)

A1 Both answers correct and in radians as multiples of π   \(-\dfrac{2\pi}{15}\) and \(\dfrac{8\pi}{15}\)
Ignore EXTRA solutions outside the range \(-\pi \lt \theta \leqslant \pi\) but lose this mark for extra solutions in this range.

Notes

NB Misread

Writing equation as \(4\cos^2 x - 7\sin x - 2 = 0\) with a sign error should be marked by applying the scheme as it simplifies the solution (do not treat as misread) Max mark is 3/6

1^st M1 Using \(\cos^2 x = 1 - \sin^2 x\) on the given equation. [Applying \(\cos^2 x = \sin^2 x - 1\), scores M0.]

1^st A1 Obtaining a correct three term equation eg. either \(4\sin^2 x - 7\sin x - 2\ \{= 0\}\) or \(-4\sin^2 x + 7\sin x + 2\ \{= 0\}\) or \(4\sin^2 x - 7\sin x = 2\) or \(4\sin^2 x = 7\sin x + 2\), etc.

2^nd M1 For a valid attempt at solving a 3TQ quadratic in sine. Methods include factorization, quadratic formula, completion of the square (unlikely here) and calculator. (See notes on page 6 for general principles on awarding this mark) Can use any variable here, \(s\), \(y\), \(x\) or \(\sin x\), and an attempt to find at least one of the solutions for \(\sin x\). This solution may be outside the range for \(\sin x\)

2^nd A1 \(\sin x = -\dfrac{1}{4}\) BY A CORRECT SOLUTION ONLY UP TO THIS POINT. Ignore extra answer of \(\sin x = 2\), but penalise if candidate states an incorrect result. e.g. \(\sin x = -2\).

Note \(\sin x = -\dfrac{1}{4}\) can be implied by later correct working if no errors are seen.

3^rd A1ft At least one of awrt 194.5 or awrt 345.5 or awrt 3.4 or awrt 6.0. This is a limited follow through. Only follow through on the error \(\sin x = \dfrac{1}{4}\) and allow for 165.5 special case (as this is equivalent work) This error is likely to earn M1A1M1A0A1A0 so 4/6 or M1A0M1A0A1A0 if the quadratic had a sign slip.

4^th A1 awrt 194.5 and awrt 345.5

Note If there are any EXTRA solutions inside the range \(0 \leqslant x \lt 360^\circ\) and the candidate would otherwise score FULL MARKS then withhold the final A1 mark.
Ignore EXTRA solutions outside the range \(0 \leqslant x \lt 360^\circ\).

Special Cases Rounding error Allow M1A1M1A1A1A0 for those who give two correct answers but wrong accuracy e.g. awrt 194, 346 (Remove final A1 for this error)
Answers in radians:– lose final mark so either or both of 3.4, 6.0 gets A1ftA0
It is possible to earn M1A0A1A1 on the final 4 marks if an error results fortuitously in \(\sin x = -1/4\) then correct work follows.