22M.1.SL.TZ2.5

determines $\frac{\mathrm{\pi }}{4}$ (or $45°$) as the first quadrant (reference) angle           (A1)

attempts to solve $\frac{x}{2}+\frac{\mathrm{\pi }}{3}=\frac{\mathrm{\pi }}{4}$           (M1)

 

Note: Award M1 for attempting to solve $\frac{x}{2}+\frac{\mathrm{\pi }}{3}=\frac{\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\left(,\dots \right)$

 

$\frac{x}{2}+\frac{\mathrm{\pi }}{3}=\frac{\mathrm{\pi }}{4}\Rightarrow x<0$ and so $\frac{\mathrm{\pi }}{4}$ is rejected           (R1)

$\frac{x}{2}+\frac{\mathrm{\pi }}{3}=2\mathrm{\pi }-\frac{\mathrm{\pi }}{4} \left(=\frac{7\mathrm{\pi }}{4}\right)$           A1

$x=\frac{17\mathrm{\pi }}{6}$  (must be in radians)           A1

 

[5 marks]

This question proved to be a struggle for many candidates, and some candidates made no attempt here. While a good number of candidates recognized the reference angle of $\frac{\pi }{4}$, this led to a final answer of $x=-\frac{\pi }{6}$, which many left as their final answer. In other cases, some candidates heeded the requirement that $x$ must be a positive value, however they gave an incorrect final answer of $x=\frac{11\pi }{6}$. Few candidates correctly rejected their initial reference angle of $\frac{\pi }{4}$ and correctly solved an equation using $\frac{x}{2}+\frac{\pi }{3}=\frac{7\pi }{4}$.