Solve the equation $2\cos x = \sin 2x$ , for $0 \le x \le 3\pi$ .

METHOD 1

using double-angle identity (seen anywhere)     A1

e.g. $\sin 2x = 2\sin x\cos x$ , $2\cos x = 2\sin x\cos x$

evidence of valid attempt to solve equation     (M1)

e.g. $0 = 2\sin x\cos x - 2\cos x$ , $2\cos x(1 - \sin x) = 0$

$\cos x = 0$ , $\sin x = 1$     A1A1

$x = \frac{\pi }{2}$ , $x = \frac{{3\pi }}{2}$ , $x = \frac{{5\pi }}{2}$     A1A1A1     N4

METHOD 2

     A1A1M1A1

Notes: Award A1 for sketch of $\sin 2x$ , A1 for a sketch of $2\cos x$ , M1 for at least one intersection point seen, and A1 for 3 approximately correct intersection points. Accept sketches drawn outside $\left[ {0,3\pi } \right]$ , even those with more than 3 intersections.

$x = \frac{\pi }{2}$ , $x = \frac{{3\pi }}{2}$ , $x = \frac{{5\pi }}{2}$     A1A1A1     N4

[7 marks]

By far the most common error was to “cancel” the $\cos x$ and find only two of the three solutions. It was disappointing how few candidates solved this by setting factors equal to zero. Some candidates wrote all three answers from $\sin x = 1$ , which only earned two of the three final marks. On a brighter note, many candidates found the $\frac{{5\pi }}{2}$ , which showed an appreciation for the period of the function as well as the domain restriction. A handful of candidates cleverly sketched both graphs and used the intersections to find the three solutions.