## Question 7:

### Part (i) — Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| For $\sec x = \frac{1}{\cos x}$ | B1 | |
| $4\sin x \cos x=1 \Rightarrow 2\sin 2x=1 \Rightarrow \sin 2x=\frac{1}{2}$ | M1 | Applying $\sin 2x=2\sin x\cos x$ and proceeding to $\sin 2x=k$, $|k|\leq1$, $k\neq0$ |
| $x=\frac{1}{2}\arcsin\!\left(\frac{1}{2}\right)$ or $\frac{1}{2}\!\left(\pi-\arcsin\!\left(\frac{1}{2}\right)\right)$ | dM1 | Correct order of operations to find at least one value |
| $x=\frac{\pi}{12}, \frac{5\pi}{12}$ | A1 | Both values and no others in range $0\leq x<\frac{\pi}{2}$ |

### Part (i) — Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| For $\sec x = \frac{1}{\cos x}$ | B1 | |
| Squaring, applying $\cos^2x+\sin^2x=1$, solving quadratic in $\sin^2x$ or $\cos^2x$: $\sin^2x$ or $\cos^2x=\frac{16\pm\sqrt{192}}{32}=\frac{2\pm\sqrt{3}}{4}$ | M1 | |
| $x=\arcsin\!\left(\sqrt{\frac{2\pm\sqrt{3}}{4}}\right)$ or $x=\arccos\!\left(\sqrt{\frac{2\pm\sqrt{3}}{4}}\right)$ | dM1 | |
| $x=\frac{\pi}{12}, \frac{5\pi}{12}$ | A1 | |

**Notes:** Give dM1 for $\sin2x=\frac{1}{2}\Rightarrow$ any of $\frac{\pi}{12}, \frac{5\pi}{12}$, 15°, 75°, awrt 0.26 or awrt 1.3; SC B1M0M0A0 for writing down $\frac{\pi}{12}, \frac{5\pi}{12}$, 15° or 75° with no working.

### Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete strategy: expresses $5\sin\theta-5\cos\theta=2$ in form $R\sin(\theta-\alpha)=2$, finds $R$ and $\alpha$, proceeds to $\sin(\theta-\alpha)=k$, $|k|<1$, $k\neq0$; **or** applies $(5\sin\theta-5\cos\theta)^2=2^2$ with $\cos^2\theta+\sin^2\theta=1$ and $\sin2\theta=2\sin\theta\cos\theta$ to reach $\sin2\theta=k$ | M1 | 3.1a |
| $R=\sqrt{50}$, $\tan\alpha=1\Rightarrow\alpha=45°$ **or** $25-25\sin2\theta=4\Rightarrow\sin2\theta=\frac{21}{25}$ | M1 | 1.1b |
| $\sin(\theta-45°)=\frac{2}{\sqrt{50}}$ **or** $\sin2\theta=\frac{21}{25}$ | A1 | 1.1b |
| e.g. $\theta=\arcsin\!\left(\frac{2}{\sqrt{50}}\right)+45°$ **or** $\theta=\frac{1}{2}\arcsin\!\left(\frac{21}{25}\right)$ | dM1 | dependent on first M mark |
| $\theta=$ awrt $61.4°$, awrt $208.6°$ | A1 | 2.1 |

**Note:** Working in radians does not affect any of the first 4 marks.

### Part (ii) — Alt 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $(5\sin\theta)^2=(2+5\cos\theta)^2$ or $(5\sin\theta-2)^2=(5\cos\theta)^2$, applies $\cos^2\theta+\sin^2\theta=1$, solves quadratic in $\sin\theta$ or $\cos\theta$ | M1 | 3.1a |
| e.g. $50\cos^2\theta+20\cos\theta-21=0$ or $50\sin^2\theta-20\sin\theta-21=0$; $\cos\theta=\frac{-20\pm\sqrt{4600}}{100}$ or $\sin\theta=\frac{20\pm\sqrt{4600}}{100}$ | M1, A1 | 1.1b |
| e.g. $\theta=\arccos\!\left(\frac{-2+\sqrt{46}}{10}\right)$ or $\theta=\arcsin\!\left(\frac{2+\sqrt{46}}{10}\right)$ | dM1 | 1.1b |
| $\theta=$ awrt $61.4°$, awrt $208.6°$ | A1 | 2.1 |

# Question 7(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| See scheme | M1 | Alternative: Express $5\sin\theta - 5\cos\theta = 2$ in form $R\cos(\theta+\alpha)=-2$, finds both $R$ and $\alpha$, proceeds to $\cos(\theta+\alpha)=k$, $|k|<1$, $k\neq 0$ |
| Uses $R\sin(\theta-\alpha)$ to find values of both $R$ and $\alpha$; or attempts $(5\sin\theta - 5\cos\theta)^2 = 2^2$, uses $\cos^2\theta+\sin^2\theta=1$ to find equation $\pm\lambda \pm \mu\sin 2\theta = \pm\beta$; or attempts $(5\sin\theta)^2=(2+5\cos\theta)^2$ and uses $\cos^2\theta+\sin^2\theta=1$ | M1 | |
| $\sin(\theta-45°)=\dfrac{2}{\sqrt{50}}$ or $\cos(\theta+45°)=-\dfrac{2}{\sqrt{50}}$ or $\sin 2\theta=\dfrac{21}{25}$ or $\cos\theta=\dfrac{-20\pm\sqrt{4600}}{100}$ giving $\cos\theta \approx 0.48$, $\approx -0.88$ | A1 | $\sin(\theta-45°)$, $\cos(\theta+45°)$, $\sin 2\theta$ must be made the subject |
| Uses correct order of operations to find at least one value of $x$ in degrees or radians; e.g. $\theta = 180°-\arcsin\!\left(\dfrac{2}{\sqrt{50}}\right)+45°$ or $\theta=\dfrac{1}{2}\!\left(180°-\arcsin\!\left(\dfrac{21}{25}\right)\right)$ | dM1 | Dependent on first M mark |
| $\theta =$ awrt $61.4°$, awrt $208.6°$ and no other values in $0\leq\theta<360°$ | A1 | Give M0M0A0M0A0 for writing down answers with no working |

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