Standard +0.3 This requires multiplying through by cos θ to form a quadratic in cos θ, then solving and checking validity of solutions. It's slightly above average due to the rational expression manipulation and domain restrictions, but follows a standard technique for this question type with straightforward algebra.
Forming a 3-term quadratic expression with all terms on the same side or correctly set up prior to completing the square. Allow \(\pm\) sign errors.
\((2\cos\theta - 1)(\cos\theta - 3) = 0\)
DM1
Solving *their* 3-term quadratic using factorisation, formula or completing the square.
\([\cos\theta = \frac{1}{2}\) or \(\cos\theta = 3\) leading to\(]\) \(\theta = -60°\) or \(\theta = 60°\)
A1
\(\theta = -60°\) and \(\theta = 60°\)
A1 FT
FT for \(\pm\) same answer between \(0°\) and \(90°\) or \(0\) and \(\frac{\pi}{2}\). \(\pm\frac{\pi}{3}\) or \(\pm 1.05\) AWRT scores maximum M1M1A0A1FT. Special case: If M1 DM0 scored then SC B1 for \(\theta = -60°\) or \(\theta = 60°\), and SC B1 FT can be awarded for \(\pm(\textit{their } 60°)\).
4
**Question 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\cos^2\theta - 7\cos\theta + 3 [= 0]$ | **M1** | Forming a 3-term quadratic expression with all terms on the same side or correctly set up prior to completing the square. Allow $\pm$ sign errors. |
| $(2\cos\theta - 1)(\cos\theta - 3) = 0$ | **DM1** | Solving *their* 3-term quadratic using factorisation, formula or completing the square. |
| $[\cos\theta = \frac{1}{2}$ or $\cos\theta = 3$ leading to$]$ $\theta = -60°$ or $\theta = 60°$ | **A1** | |
| $\theta = -60°$ **and** $\theta = 60°$ | **A1 FT** | FT for $\pm$ same answer between $0°$ and $90°$ or $0$ and $\frac{\pi}{2}$. $\pm\frac{\pi}{3}$ or $\pm 1.05$ AWRT scores maximum M1M1A0A1FT. **Special case:** If M1 DM0 scored then SC B1 for $\theta = -60°$ **or** $\theta = 60°$, and SC B1 FT can be awarded for $\pm(\textit{their } 60°)$. |
| | **4** | |