## Question 5:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses identity $\cot^2(3\theta) = \csc^2(3\theta)-1$ in $2\cot^2(3\theta) = 7\csc(3\theta)-5$ | M1 | Accept 'invisible' brackets; longer alternative is to convert to $\sin(3\theta)$; $\cot^2 3\theta$ replaced by $\frac{\cos^2(3\theta)}{\sin^2(3\theta)}$, $\csc(3\theta)$ by $\frac{1}{\sin 3\theta}$, then multiply by $\sin^2(3\theta)$ |
| $2\csc^2(3\theta) - 7\csc(3\theta) + 3 = 0$ | A1 | Correct equation; terms collected on one side |
| $(2\csc 3\theta - 1)(\csc 3\theta - 3) = 0$ | dM1 | Attempt to factorise 3-term quadratic in $\csc(3\theta)$ or $\sin(3\theta)$, or use correct formula |
| $\csc 3\theta = 3$ | A1 | Correct value; ignore other values (note $\csc 3\theta = \frac{1}{2}$ is impossible) |
| $\theta = \frac{\arcsin(\frac{1}{3})}{3}$, $\approx \frac{19.5°}{3}$ = awrt $6.5°$ | ddM1, A1 | Correct method for principal value; dependent on two M marks |
| $\theta = \frac{180° - \arcsin(\frac{1}{3})}{3}$, $53.5°$ | ddM1, A1 | Correct 2nd value; look for $\frac{180°-\text{their }19.5°}{3}$; **Note: $\frac{360°+\text{their }6.5°}{3}$ is incorrect** |
| $\theta = \frac{360° + \arcsin(\frac{1}{3})}{3}$ | ddM1 | Correct 3rd value; dependent on first 2 M marks |
| All 4 correct answers awrt $6.5°, 53.5°, 126.5°, 173.5°$ | A1 | No extra answers inside range; ignore answers outside range |

**Additional guidance:** Radian answers awrt $0.11, 0.93, 2.21, 3.03$ (accuracy to 2dp). Candidates cannot mix degrees and radians for method marks.