Standard +0.3 Part (i) requires algebraic manipulation to rearrange a rational equation and solve a linear equation in sin 2θ, then find angles in a restricted range. Part (ii) requires the standard identity cos²x + sin²x = 1 to convert to a quadratic in cos x, then solve and find multiple solutions. Both are routine C2-level techniques with straightforward application of identities and solving methods, slightly above average due to the algebraic manipulation and multiple solutions required.
7. (i) Solve, for \(0 \leqslant \theta < 180 ^ { \circ }\), the equation
$$\frac { \sin 2 \theta } { ( 4 \sin 2 \theta - 1 ) } = 1$$
giving your answers to 1 decimal place.
(ii) Solve, for \(0 \leqslant x < 2 \pi\), the equation
$$5 \sin ^ { 2 } x - 2 \cos x - 5 = 0$$
giving your answers to 2 decimal places. (Solutions based entirely on graphical or numerical methods are not acceptable.)
\(\sin 2\theta = k\) where \(-1 < k < 1\). Must be \(2\theta\) and not \(\theta\)
\(\{2\theta = \{19.4712..., 160.5288...\}\}\)
\(\theta = \{9.7356..., 80.2644...\}\)
A1 A1
A1: Either awrt 9.7 or awrt 80.3. A1: Both awrt 9.7 and awrt 80.3
> Do not penalise poor accuracy more than once e.g. 9.8 and 80.2 from correct work could score M1A1A0. If both answers are correct in radians award A1A0 otherwise A0A0. Correct answers are 0.2 and 1.4. Extra solutions in range in an otherwise fully correct solution deduct the last A1.
Part (ii):
\[5\sin^2 x - 2\cos x - 5 = 0,\ 0\leq x < 2\pi\]
Answer
Marks
Guidance
Answer
Mark
Guidance
\(5(1-\cos^2 x) - 2\cos x - 5 = 0\)
M1
Applies \(\sin^2 x = 1 - \cos^2 x\)
\(5\cos^2 x + 2\cos x = 0\), \(\cos x(5\cos x + 2) = 0 \Rightarrow \cos x = \ldots\)
dM1
Cancelling out \(\cos x\) or a valid attempt at solving the quadratic in \(\cos x\) and giving \(\cos x = \ldots\) Dependent on previous method mark
awrt 1.98 or awrt 4.3(0)
A1
Degrees: 113.58, 246.42
Both 1.98 and 4.3(0)
A1ft
or their \(\alpha\) and their \(2\pi - \alpha\), where \(\alpha \neq \frac{\pi}{2}\). If working in degrees allow \(360 -\) their \(\alpha\)
awrt 1.57 or \(\frac{\pi}{2}\) and 4.71 or \(\frac{3\pi}{2}\), or \(90°\) and \(270°\)
B1
These answers only but ignore other answers outside the range
# Question 7:
## Part (i):
$$\frac{\sin 2\theta}{(4\sin 2\theta - 1)} = 1;\ 0\leqslant\theta < 180°$$
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sin 2\theta = \frac{1}{3}$ | M1 | $\sin 2\theta = k$ where $-1 < k < 1$. **Must be $2\theta$ and not $\theta$** |
| $\{2\theta = \{19.4712..., 160.5288...\}\}$ | | |
| $\theta = \{9.7356..., 80.2644...\}$ | A1 A1 | A1: Either awrt 9.7 or awrt 80.3. A1: Both awrt 9.7 and awrt 80.3 |
> Do not penalise poor accuracy more than once e.g. 9.8 and 80.2 from correct work could score M1A1A0. If **both** answers are correct in radians award A1A0 otherwise A0A0. Correct answers are 0.2 and 1.4. Extra solutions in range in an otherwise fully correct solution deduct the last A1.
## Part (ii):
$$5\sin^2 x - 2\cos x - 5 = 0,\ 0\leq x < 2\pi$$
| Answer | Mark | Guidance |
|--------|------|----------|
| $5(1-\cos^2 x) - 2\cos x - 5 = 0$ | M1 | Applies $\sin^2 x = 1 - \cos^2 x$ |
| $5\cos^2 x + 2\cos x = 0$, $\cos x(5\cos x + 2) = 0 \Rightarrow \cos x = \ldots$ | dM1 | Cancelling out $\cos x$ or a valid attempt at solving the quadratic in $\cos x$ **and** giving $\cos x = \ldots$ Dependent on previous method mark |
| awrt 1.98 or awrt 4.3(0) | A1 | Degrees: 113.58, 246.42 |
| Both 1.98 and 4.3(0) | A1ft | or their $\alpha$ and their $2\pi - \alpha$, where $\alpha \neq \frac{\pi}{2}$. If working in degrees allow $360 -$ their $\alpha$ |
| awrt 1.57 or $\frac{\pi}{2}$ **and** 4.71 or $\frac{3\pi}{2}$, or $90°$ and $270°$ | B1 | These answers only but ignore other answers **outside** the range |
> NB: $x =$ awrt $\left\{1.98, 4.3(0), 1.57\ \text{or}\ \frac{\pi}{2}, 4.71\ \text{or}\ \frac{3\pi}{2}\right\}$. Answers in degrees: 113.58, 246.42, 90, 270. Could score M1M1A0A1ftB1 (4/5).
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