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.