Solve \(\cos 2x - 3\cos x - 3 - {\cos ^2}x = {\sin ^2}x\) , for \(0 \le x \le 2\pi \) .

evidence of substituting for \(\cos 2x\)     (M1)

evidence of substituting into \({\sin ^2}x + {\cos ^2}x = 1\)     (M1)

correct equation in terms of \(\cos x\) (seen anywhere)     A1

e.g. \(2{\cos ^2}x - 1 - 3\cos x - 3 = 1\) , \(2{\cos ^2}x - 3\cos x - 5 = 0\)

evidence of appropriate approach to solve     (M1)

e.g. factorizing, quadratic formula

appropriate working     A1

e.g. \((2\cos x - 5)(\cos x + 1) = 0\) , \((2x - 5)(x + 1)\) , \(\cos x = \frac{{3 \pm \sqrt {49}}}{4}\)

correct solutions to the equation

e.g. \(\cos x = \frac{5}{2}\) , \(\cos x = - 1\) , \(x = \frac{5}{2}\) , \(x = - 1\)     (A1)

\(x = \pi \)     A1     N4

[7 marks]

This question was quite difficult for most candidates. A number of students earned some credit for manipulating the equation with identities, but many earned no further marks due to algebraic errors. Many did not substitute for \(\cos 2x\) ; others did this substitution but then did nothing further.

Few candidates were able to get a correct equation in terms of \(\cos x\) and many who did get the equation didn't know what to do with it. Candidates who correctly solved the resulting quadratic usually found the one correct value of \(x\), earning full marks.