Show that $4 - \cos 2\theta + 5\sin \theta = 2{\sin ^2}\theta + 5\sin \theta + 3$ .

Hence, solve the equation $4 - \cos 2\theta + 5\sin \theta = 0$ for $0 \le \theta \le 2\pi$ .

attempt to substitute $1 - 2{\sin ^2}\theta$ for $\cos 2\theta$     (M1)

correct substitution     A1

e.g. $4 - (1 - 2{\sin ^2}\theta ) + 5\sin\theta$

$4 - \cos 2\theta + 5\sin \theta = 2{\sin ^2}\theta + 5\sin\theta + 3$     AG     N0

[2 marks]

evidence of appropriate approach to solve     (M1)

e.g. factorizing, quadratic formula

correct working     A1

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

correct solution $\sin \theta = - 1$ (do not penalise for including $\sin \theta = - \frac{3}{2}$     (A1)

$\theta = \frac{{3\pi }}{2}$     A2     N3

[5 marks]

In part (a), most candidates successfully substituted using the double-angle formula for cosine. There were quite a few candidates who worked backward, starting with the required answer and manipulating the equation in various ways. As this was a "show that" question, working backward from the given answer is not a valid method.

In part (b), many candidates seemed to realize what was required by the word “hence”, though some had trouble factoring the quadratic-type equation. A few candidates were also successful using the quadratic formula. Some candidates got the wrong solution to the equation $\sin \theta = - 1$ , and there were a few who did not realize that the equation $\sin \theta = - \frac{3}{2}$ has no solution.