# Question 1:

## Part (a)

| Working/Answer | Marks | Guidance |
|---|---|---|
| $5\sin x + 12\cos x = 5\left(\frac{2t}{1+t^2}\right) + 12\left(\frac{1-t^2}{1+t^2}\right)$ | M1 | Uses at least one of $\sin x = \frac{2t}{1+t^2}$ or $\cos x = \frac{1-t^2}{1+t^2}$ to express $5\sin x + 12\cos x$ in terms of $t$ only |
| $5\left(\frac{2t}{1+t^2}\right) + 12\left(\frac{1-t^2}{1+t^2}\right) = 2 \Rightarrow 5(2t) + 12(1-t^2) = 2(1+t^2)$ | M1 | Uses both correct formulae, equates to 2 and eliminates fractions |
| $7t^2 - 5t - 5 = 0$ | A1* | Collects terms to one side and simplifies to obtain printed answer |

**(3 marks)**

## Part (b)

| Working/Answer | Marks | Guidance |
|---|---|---|
| $t = \frac{5 \pm \sqrt{5^2 - 4(7)(-5)}}{2(7)} \left\{= \frac{5 \pm \sqrt{165}}{14} = 1.2746..., -0.5603...\right\}$ | M1 | Selects correct process (quadratic formula, completing the square or calculator) to solve $7t^2 - 5t - 5 = 0$. Note: factorisation attempt not allowed for 1st M1 |
| $\frac{x}{2} = \arctan\left(\frac{5+\sqrt{165}}{14}\right)$ or $\frac{x}{2} = \arctan\left(\frac{5-\sqrt{165}}{14}\right)$ and $\Rightarrow x = ...$ | M1 | Adopts correct **applied** strategy of taking $\arctan$(their found $t$) and multiplying by 2 to obtain at least one value for $x$ within range $-180° < x < 180°$ |
| $x =$ awrt $104°$ **or** $x =$ any answer in range $[-58.6°, -58°]$ | A1 | See scheme. Give final A0 for extra solutions within range $-180° < x < 180°$ |
| $x = 103.8°$ **and** $x = -58.5°$ | A1 | For both 103.8 and $-58.5$. Ignore extra solutions outside range for final A mark |

**(4 marks)**

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