If x satisfies the equation $\sin \left( {x + \frac{\pi }{3}} \right) = 2\sin x\sin \left( {\frac{\pi }{3}} \right)$, show that $11\tan x = a + b\sqrt 3$, where a, b $\in {\mathbb{Z}^ + }$.

$\sin \left( {x + \frac{\pi }{3}} \right) = \sin x\cos \left( {\frac{\pi }{3}} \right) + \cos x\sin \left( {\frac{\pi }{3}} \right)$     (M1)

$\sin x\cos \left( {\frac{\pi }{3}} \right) + \cos x\sin \left( {\frac{\pi }{3}} \right) = 2\sin x\sin \left( {\frac{\pi }{3}} \right)$

$\frac{1}{2}\sin x + \frac{{\sqrt 3 }}{2}\cos x = 2 \times \frac{{\sqrt 3 }}{2}\sin x$     A1

dividing by $\cos x$ and rearranging     M1

$\tan x = \frac{{\sqrt 3 }}{{2\sqrt 3 - 1}}$     A1

rationalizing the denominator     M1

$11\tan x = 6 + \sqrt 3$     A1

[6 marks]

Most candidates were able to make a meaningful start to this question, but a significant number were unable to find an appropriate expression for $\tan x$ or to rationalise the denominator.