| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show equivalence then solve inequality |
| Difficulty | Standard +0.3 Part (a) is a straightforward algebraic manipulation using sin²θ + cos²θ = 1 and tanθ = sinθ/cosθ. Part (b) requires solving a quadratic in sinθ and interpreting solutions graphically for an inequality. Standard techniques throughout with no novel insight required, but slightly above average due to the inequality interpretation step. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(6(1-\sin^2\theta) = \dfrac{\sin\theta}{\cos\theta}(\cos\theta) + 4\) | M1 | Correct use of both \(\cos^2\theta = 1 - \sin^2\theta\) and \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\) |
| \(6 - 6\sin^2\theta = \sin\theta + 4 \Rightarrow 6\sin^2\theta + \sin\theta - 2 = 0\) | A1 | AG; must show sufficient working to justify the given answer |
| Answer | Marks | Guidance |
|---|---|---|
| \((2\sin\theta - 1)(3\sin\theta + 2)\) | M1 | Attempt to solve 3-term quadratic; ignore incorrect use of inequalities for first three marks |
| Critical values when \(\sin\theta = \dfrac{1}{2}\) and \(\sin\theta = -\dfrac{2}{3}\) | B1 | |
| Critical values are \(\theta = 30, 150, 222, 318\) | B1 | Any three correct critical values |
| \(0 < \theta < 30\) or \(150 < \theta < 222\) or \(318 < \theta < 360\) | B1 | B1 for one correct interval – 3sf or better |
| A1 | Cao (all three intervals) – 3sf or better; allow \(\theta < 30\), \(150 < \theta < 222\), \(318 < \theta\) |
# Question 6:
## Part (a):
$6(1-\sin^2\theta) = \dfrac{\sin\theta}{\cos\theta}(\cos\theta) + 4$ | **M1** | Correct use of both $\cos^2\theta = 1 - \sin^2\theta$ and $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$
$6 - 6\sin^2\theta = \sin\theta + 4 \Rightarrow 6\sin^2\theta + \sin\theta - 2 = 0$ | **A1** | AG; must show sufficient working to justify the given answer
## Part (b):
$(2\sin\theta - 1)(3\sin\theta + 2)$ | **M1** | Attempt to solve 3-term quadratic; ignore incorrect use of inequalities for first three marks
Critical values when $\sin\theta = \dfrac{1}{2}$ and $\sin\theta = -\dfrac{2}{3}$ | **B1** |
Critical values are $\theta = 30, 150, 222, 318$ | **B1** | Any three correct critical values
$0 < \theta < 30$ or $150 < \theta < 222$ or $318 < \theta < 360$ | **B1** | B1 for one correct interval – 3sf or better
| **A1** | Cao (all three intervals) – 3sf or better; allow $\theta < 30$, $150 < \theta < 222$, $318 < \theta$
---\begin{enumerate}[label=(\alph*)]
\item Show that the equation $6 \cos ^ { 2 } \theta = \tan \theta \cos \theta + 4$\\
can be expressed in the form $6 \sin ^ { 2 } \theta + \sin \theta - 2 = 0$.
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{d6430776-0b87-4e5e-8f78-c6228ee163d5-4_446_1150_1119_338}
The diagram shows parts of the curves $y = 6 \cos ^ { 2 } \theta$ and $y = \tan \theta \cos \theta + 4$, where $\theta$ is in degrees.
Solve the inequality $6 \cos ^ { 2 } \theta > \tan \theta \cos \theta + 4$ for $0 ^ { \circ } < \theta < 360 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q6 [7]}}