verify that $x = \frac{\pi }{{12}}$ is a solution to the equation;

hence find the other solution to the equation for $0 < x < \frac{\pi }{2}$.

EITHER

${\text{LHS}} = \frac{{\sqrt 3  - 1}}{{\frac{{\sqrt 6  - \sqrt 2 }}{4}}} + \frac{{\sqrt 3  + 1}}{{\frac{{\sqrt 6  + \sqrt 2 }}{4}}}$    M1

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

$= 2\sqrt 2  + 2\sqrt 2$    A1

${\text{LHS}} = 4\sqrt 2  \Rightarrow x = \frac{\pi }{{12}}$ is a solution     AG

OR

${\text{LHS}} = \frac{{\sqrt 3  - 1}}{{\frac{{\sqrt 6  - \sqrt 2 }}{4}}} + \frac{{\sqrt 3  + 1}}{{\frac{{\sqrt 6  + \sqrt 2 }}{4}}}$    M1

$= \frac{{\left( {\sqrt 3  - 1} \right)\left( {\frac{{\sqrt 6  + \sqrt 2 }}{4}} \right) + \left( {\sqrt 3  + 1} \right)\left( {\frac{{\sqrt 6  - \sqrt 2 }}{4}} \right)}}{{\left( {\frac{{\sqrt 6  - \sqrt 2 }}{4}} \right)\left( {\frac{{\sqrt 6  + \sqrt 2 }}{4}} \right)}}$    A1

$= 2\sqrt {18}  - 2\sqrt 2$ (or equivalent)     A1

${\text{LHS}} = 4\sqrt 2  \Rightarrow x = \frac{\pi }{{12}}$ is a solution     AG

[3 marks]

$\frac{{\sqrt 2 }}{4}\left( {\frac{{\sqrt 3  - 1}}{{\sin x}} + \frac{{\sqrt 3  + 1}}{{\cos x}}} \right) = 2 \Rightarrow \frac{{\sin \frac{\pi }{{12}}}}{{\sin x}} + \frac{{\cos \frac{\pi }{{12}}}}{{\cos x}} = 2$    M1

$\frac{{\sin \frac{\pi }{{12}}\cos x + \cos \frac{\pi }{{12}}\sin x}}{{\sin x\cos x}} = 2$    M1

$\sin \frac{\pi }{{12}}\cos x + \cos \frac{\pi }{{12}}\sin x = 2\sin x\cos x$

$\sin \left( {\frac{\pi }{{12}} + x} \right) = \sin 2x$    A1

$\frac{\pi }{{12}} + x = \pi  - 2x$ or $\pi  - \left( {\frac{\pi }{{12}} + x} \right) = 2x$     (M1)

$x = \frac{{11\pi }}{{36}}$    A1

[5 marks]

This question proved to be the most problematic question in the paper.

Part (a) was generally well done, with competent fraction and surd manipulation seen successfully in leading to the given answer.

This question proved to be the most problematic question in the paper.

The number of scripts seen where part (b) was tackled with complete success numbered in the single figures; solutions were rarely if ever seen. Some candidates scored one mark by finding, or using, the common denominator $\sin x\cos x$.