Moderate -0.3 This is a straightforward double angle equation requiring the standard substitution cos 2θ = 2cos²θ - 1, leading to a quadratic in cos θ. The algebraic manipulation is routine and finding solutions in the given range is standard practice. Slightly easier than average due to being a direct application of a common technique with no conceptual surprises.
M1 A0 if they use any correct formula for \(\cos 2\theta\) and make a subsequent error
Solve a 3-term quadratic in \(\cos\theta\) for \(\theta\)
M1
As far as \(\theta = \cos^{-1}\left(\frac{3-\sqrt{129}}{12}\right)\) if quadratic correct
Obtain a correct answer, e.g. \(134.1°\)
A1
Accept greater accuracy e.g. 134.1456, 225.8544
Obtain a second answer, e.g. \(225.9°\) and no other in \([0°, 360°]\)
A1 FT
Treat answers in radians (2.34 and 3.94) as misread. Ignore answers outside \([0°, 360°]\). FT is for \(360°\) minus the first answer. Special Ruling: If they have an incorrect quadratic leading legitimately to 4 solutions, allow FT for \(360°\) minus an answer in \((0°, 180°)\). More than 4 solutions is maximum M1 A0 M1 A0 A0.
Total: 5 marks
## Question 2:
| Use correct double-angle formula to obtain an equation in $\cos\theta$ | M1 | e.g. $3(2\cos^2\theta - 1) = 3\cos\theta + 2$ |
|---|---|---|
| Obtain $6\cos^2\theta - 3\cos\theta - 5 = 0$, or 3-term equivalent | A1 | M1 A0 if they use any correct formula for $\cos 2\theta$ and make a subsequent error |
| Solve a 3-term quadratic in $\cos\theta$ for $\theta$ | M1 | As far as $\theta = \cos^{-1}\left(\frac{3-\sqrt{129}}{12}\right)$ if quadratic correct |
| Obtain a correct answer, e.g. $134.1°$ | A1 | Accept greater accuracy e.g. 134.1456, 225.8544 |
| Obtain a second answer, e.g. $225.9°$ and no other in $[0°, 360°]$ | A1 FT | Treat answers in radians (2.34 and 3.94) as misread. Ignore answers outside $[0°, 360°]$. FT is for $360°$ minus the first answer. Special Ruling: If they have an incorrect quadratic leading legitimately to 4 solutions, allow FT for $360°$ minus an answer in $(0°, 180°)$. More than 4 solutions is maximum M1 A0 M1 A0 A0. |
**Total: 5 marks**
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