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.