| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Reduce to quadratic in trig |
| Difficulty | Moderate -0.3 Part (a) is a one-step linear equation in tan x requiring basic inverse tan and periodicity. Part (b) requires using the Pythagorean identity to convert to a quadratic in sin x, then solving—a standard technique but with multiple steps. Overall slightly easier than average due to straightforward methods and no conceptual surprises. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = \tan^{-1}\!\left(\tfrac{3}{2}\right)\) | M1 | Attempt inverse tan of \(\frac{3}{2}\); may be implied by result, e.g. \(\tan x = \frac{3}{2}\) |
| \(x = 56.3°\) (3 sf) | A1 | Allow omission of degrees sign throughout |
| \(x = 236°\) (3 sf) with no extras | A1 | SC: If no working shown, B2 both correct no extras; B1 one correct no extras or both correct with extras |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| DR \(5\sin x - 1 = 2(1 - \sin^2 x)\), \(2\sin^2 x + 5\sin x - 3 = 0\) | M1 | Use of \(\sin^2 + \cos^2 = 1\). May be implied |
| \((2\sin x - 1)(\sin x + 3) = 0\) or \(\sin x = \frac{-5 \pm \sqrt{25+24}}{4}\) | M1 | or \((2u-1)(u+3)=0\) or \(u = \frac{-5\pm\sqrt{25+24}}{4}\). Correct method seen, ft their equation |
| \(\sin x = 0.5\) (or \(-3\)) or \(u = 0.5\) (or \(-3\)) | A1f | ft their equation. Allow without \(\sin x = -3\). Dep 1st M1, not 2nd M1 |
| \(\sin x = -3\) is not possible, or no solution oe | B1 | Appropriate comment needed, e.g. "N/A", not just crossing out |
| \(x = 30°\) or \(150°\) | A1 | cao. Both, with no extras. Dep 1st M1, not 2nd M1 |
| [5] |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \tan^{-1}\!\left(\tfrac{3}{2}\right)$ | M1 | Attempt inverse tan of $\frac{3}{2}$; may be implied by result, e.g. $\tan x = \frac{3}{2}$ |
| $x = 56.3°$ (3 sf) | A1 | Allow omission of degrees sign throughout |
| $x = 236°$ (3 sf) with no extras | A1 | SC: If no working shown, B2 both correct no extras; B1 one correct no extras or both correct with extras |
| **[3]** | | |
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## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **DR** $5\sin x - 1 = 2(1 - \sin^2 x)$, $2\sin^2 x + 5\sin x - 3 = 0$ | M1 | Use of $\sin^2 + \cos^2 = 1$. May be implied |
| $(2\sin x - 1)(\sin x + 3) = 0$ or $\sin x = \frac{-5 \pm \sqrt{25+24}}{4}$ | M1 | or $(2u-1)(u+3)=0$ or $u = \frac{-5\pm\sqrt{25+24}}{4}$. Correct method **seen**, ft their equation |
| $\sin x = 0.5$ (or $-3$) or $u = 0.5$ (or $-3$) | A1f | ft their equation. Allow without $\sin x = -3$. Dep 1st M1, not 2nd M1 |
| $\sin x = -3$ is not possible, or no solution oe | B1 | Appropriate comment needed, e.g. "N/A", not just crossing out |
| $x = 30°$ or $150°$ | A1 | cao. Both, with no extras. Dep 1st M1, not 2nd M1 |
| **[5]** | | |
---4 In this question you must show detailed reasoning.
Solve the following equations, for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\begin{enumerate}[label=(\alph*)]
\item $2 \tan x + 1 = 4$
\item $5 \sin x - 1 = 2 \cos ^ { 2 } x$
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q4 [8]}}