## Question 9(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin(3x-15) = \frac{1}{2}$, so $3x - 15 = 30$ $(\alpha)$, $x = 15$ | M1, A1 | Correct order: inverse sine then linear algebra |
| $3x - 15 = 180 - \alpha$ or $3x - 15 = 540 - \alpha$ | M1 | Uses either $180-\alpha$ or $540-\alpha$ |
| $3x - 15 = 180-\alpha$ and $3x-15 = 360+\alpha$ and $3x-15 = 540-\alpha$ | M1 | Uses all three forms |
| $x = 55$ or $175$ | A1 | One further correct solution |
| $x = 55, 135, 175$ | A1 | All three correct; more than 4 solutions loses last A1 |

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## Question 9(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| At least one of $\left(\frac{a\pi}{10} - b\right) = 0$ or $\left(\frac{a3\pi}{5} - b\right) = \pi$ or $\left(\frac{a11\pi}{10} - b\right) = 2\pi$ | M1 | Award for $\frac{a\pi}{10} - b = 0$ or $\frac{a\pi}{10} = b$; NOT $\sin\left(\frac{a\pi}{10}-b\right)=0$ |
| Uses two equations to eliminate $a$ or $b$, or uses period property to find $a$ | M1 | e.g. $\frac{5\pi}{4}a = \pi$ so $a=2$ |
| $a = 2$ | A1 | |
| $b = \frac{\pi}{5}$ | A1 | Must be in radians |

The image you've shared appears to be only the back cover/colophon page of an Edexcel publication from January 2012, containing only publisher information (contact details, order codes, and logos for Ofqual, Welsh Assembly Government, and CEA).

There is **no mark scheme content** visible on this page to extract.

Could you please share the actual mark scheme pages? They would typically contain question numbers, model answers, mark allocations (M1, A1, B1, etc.), and examiner guidance notes.