Moderate -0.3 This is a standard C2 trigonometric equation requiring the identity cos²x = 1 - sin²x to convert to a quadratic in sin x, then solving the quadratic and finding angles in the given range. It's slightly easier than average because the conversion and solution steps are routine, though students must remember to check which solutions are valid.
Allow 3sf or better. Must come from a correct solution of the correct quadratic.
Obtain all 3 correct solutions, and no others
A1
Must now all be in degrees. Allow 3sf or better. A0 if other incorrect solutions in range \(0°\)–\(360°\).
[6]
## Question 4:
### Solve $4\cos^2 x + 7\sin x - 7 = 0$ for $0° \leq x \leq 360°$
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4(1-\sin^2 x) + 7\sin x - 7 = 0$, use $\cos^2 x = 1 - \sin^2 x$ | M1 | Must be used and not just stated. Must be used correctly, so M0 for $1 - 4\sin^2 x$. |
| $4\sin^2 x - 7\sin x + 3 = 0$, obtain correct quadratic | A1 | aef, as long as three term quadratic with all terms on one side. Condone $4\sin^2 x - 7\sin x + 3$ ie no $= 0$. |
| $(\sin x - 1)(4\sin x - 3) = 0$, attempt to solve quadratic in $\sin x$ | M1 | Not dependent on previous M1. This M mark is just for solving a 3 term quadratic. Condone any substitution used, inc $x = \sin x$. |
| $\sin x = 1$, $\sin x = \frac{3}{4}$, attempt to find $x$ from roots of quadratic | M1 | Attempt $\sin^{-1}$ of at least one of their roots. Allow for just stating $\sin^{-1}$ (their root) inc if $|\sin x| > 1$. Not dependent on previous marks. If going straight from $\sin x = k$ to $x = \ldots$, then award M1 only if their angle is consistent with their $k$. |
| $x = 90°$, $x = 48.6°$, $131°$, obtain two correct solutions | A1 | Allow 3sf or better. Must come from a correct solution of the correct quadratic. |
| Obtain all 3 correct solutions, and no others | A1 | Must now all be in degrees. Allow 3sf or better. A0 if other incorrect solutions in range $0°$–$360°$. |
| | **[6]** | |