OCR C2 — Question 8 11 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTrigonometric equations in context
TypeSolve double/multiple angle equation
DifficultyStandard +0.3 Part (i) is straightforward application of inverse tan with double angle consideration. Part (ii) requires rearranging to sin y(2cos y - 1) = 0 and solving, which is standard C2 technique but slightly more demanding than routine exercises. Overall slightly easier than average A-level questions.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
8. (i) Find, to 2 decimal places, the values of \(x\) in the interval \(0 \leq x < \pi\) for which $$\tan 2 x = 3$$ (ii) Find, in terms of \(\pi\), the values of \(y\) in the interval \(0 \leq y < 2 \pi\) for which $$2 \sin y = \tan y$$
(i) \(2x = 1.2490, \pi + 1.2490 = 1.2490, 4.3906\)
AnswerMarks
\(x = 0.62, 2.20\) (2dp)B1 M1 M1 A1
(ii) \(2\sin y \cos y = \sin y\)
AnswerMarks Guidance
\(\sin y(2\cos y - 1) = 0\)M1 M1
\(\sin y = 0\) or \(\cos y = \frac{1}{2}\)A1
\(y = 0, \pi\) or \(y = \frac{\pi}{3}, 2\pi - \frac{\pi}{3}\)B2 M1
\(y = 0, \frac{\pi}{3}, \pi, \frac{2\pi}{3}\)A1 (11) 
(i) $2x = 1.2490, \pi + 1.2490 = 1.2490, 4.3906$

$x = 0.62, 2.20$ (2dp) | B1 M1 M1 A1 |

(ii) $2\sin y \cos y = \sin y$

$\sin y(2\cos y - 1) = 0$ | M1 M1 |

$\sin y = 0$ or $\cos y = \frac{1}{2}$ | A1 |

$y = 0, \pi$ or $y = \frac{\pi}{3}, 2\pi - \frac{\pi}{3}$ | B2 M1 |

$y = 0, \frac{\pi}{3}, \pi, \frac{2\pi}{3}$ | A1 | **(11)** |

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8. (i) Find, to 2 decimal places, the values of $x$ in the interval $0 \leq x < \pi$ for which

$$\tan 2 x = 3$$

(ii) Find, in terms of $\pi$, the values of $y$ in the interval $0 \leq y < 2 \pi$ for which

$$2 \sin y = \tan y$$

\hfill \mbox{\textit{OCR C2  Q8 [11]}}
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