For what values of x does $f(x)$ not exist?

Simplify the expression $\frac{{\sin 3x}}{{\sin x}} - \frac{{\cos 3x}}{{\cos x}}$.

$\cos x = 0{\text{, }}\sin x = 0$     (M1)

$x = \frac{{n\pi }}{2},n \in \mathbb{Z}$     A1

EITHER

$\frac{{\sin 3x\cos x - \cos 3x\sin x}}{{\sin x\cos x}}$     M1     A1

$= \frac{{\sin (3x - x)}}{{\frac{1}{2}\sin 2x}}$     A1     A1

$= 2$     A1

OR

$\frac{{\sin 2x\cos x + \cos 2x\sin x}}{{\sin x}} - \frac{{\cos 2x\cos x - \sin 2x\sin x}}{{\cos x}}$     M1

$= \frac{{2\sin x{{\cos }^2}x + 2{{\cos }^2}x\sin x - \sin x}}{{\sin x}} - \frac{{2{{\cos }^3}x - \cos x - {{\sin }^2}x\cos x}}{{\cos x}}$     A1     A1

$= 4{\cos ^2}x - 1 - 2{\cos ^2}x + 1 + 2{\sin ^2}x$     A1
$= 2{\cos ^2}x + 2{\sin ^2}x$

$= 2$     A1

[5 marks]

Part (a) was well answered, although many candidates lost a mark through not giving sufficient solutions. It was rare for a student to receive no marks for part (b), but few solved the question by the easiest route, and as a consequence, there were frequently errors in the algebraic manipulation of the expression. 

Part (a) was well answered, although many candidates lost a mark through not giving sufficient solutions. It was rare for a student to receive no marks for part (b), but few solved the question by the easiest route, and as a consequence, there were frequently errors in the algebraic manipulation of the expression.