Edexcel C2 2014 June — Question 7 9 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Year2014
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTrigonometric equations in context
TypeSolve shifted trig equation
DifficultyModerate -0.3 Part (i) is a straightforward single-step trigonometric equation requiring inverse sine and consideration of two solutions in the given range. Part (ii) requires factorising to sin x(2sec x - 3) = 0, then solving two cases, which is standard C2 technique but involves slightly more algebraic manipulation. Both parts are routine exercises with clear methods and no novel problem-solving required, making this slightly easier than average.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals

7. (i) Solve, for \(0 \leqslant \theta < 360 ^ { \circ }\), the equation $$9 \sin \left( \theta + 60 ^ { \circ } \right) = 4$$ giving your answers to 1 decimal place.
You must show each step of your working.
(ii) Solve, for \(- \pi \leqslant x < \pi\), the equation $$2 \tan x - 3 \sin x = 0$$ giving your answers to 2 decimal places where appropriate. [Solutions based entirely on graphical or numerical methods are not acceptable.]

Question 7:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\sin(\theta+60°) = \frac{4}{9}\), so \((\theta+60°) = 26.3877...\)M1 Sight of \(\sin^{-1}\left(\frac{4}{9}\right)\) or awrt 26.4° or 0.461°; can be implied by \(\theta =\) awrt \(-33.6\)
\(\theta + 60° = \{153.6122..., 386.3877...\}\)M1 \(\theta+60°=\) either "180 – their \(\alpha\)" or "360° + their \(\alpha\)" and not for \(\theta=\) either; can be implied by later working
\(\theta = \{93.6122..., 326.3877...\}\)A1 A1 A1: at least one of awrt 93.6° or awrt 326.4°; A1: both awrt 93.6° and awrt 326.4°
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(2\left(\frac{\sin x}{\cos x}\right) - 3\sin x = 0\)M1 Applies \(\tan x = \frac{\sin x}{\cos x}\)
\(\sin x(2 - 3\cos x) = 0\)
\(\cos x = \frac{2}{3}\)A1 \(\cos x = \frac{2}{3}\)
\(x = \text{awrt}\{0.84, -0.84\}\)A1 A1ft A1: one of awrt 0.84 or awrt \(-0.84\); A1ft: apply ft for \(x = \pm\alpha\) where \(\alpha = \cos^{-1}k\) and \(-1 \leq k \leq 1\)
\(\{\sin x = 0 \Rightarrow\}\ x = 0\) and \(-\pi\)B1 Both \(x=0\) and \(-\pi\) or awrt \(-3.14\) from \(\sin x = 0\); ignore extra solutions in range
## Question 7:

### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin(\theta+60°) = \frac{4}{9}$, so $(\theta+60°) = 26.3877...$ | M1 | Sight of $\sin^{-1}\left(\frac{4}{9}\right)$ or awrt 26.4° or 0.461°; can be implied by $\theta =$ awrt $-33.6$ |
| $\theta + 60° = \{153.6122..., 386.3877...\}$ | M1 | $\theta+60°=$ either "180 – their $\alpha$" or "360° + their $\alpha$" and not for $\theta=$ either; can be implied by later working |
| $\theta = \{93.6122..., 326.3877...\}$ | A1 A1 | A1: at least one of awrt 93.6° or awrt 326.4°; A1: both awrt 93.6° and awrt 326.4° |

### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\left(\frac{\sin x}{\cos x}\right) - 3\sin x = 0$ | M1 | Applies $\tan x = \frac{\sin x}{\cos x}$ |
| $\sin x(2 - 3\cos x) = 0$ | | |
| $\cos x = \frac{2}{3}$ | A1 | $\cos x = \frac{2}{3}$ |
| $x = \text{awrt}\{0.84, -0.84\}$ | A1 A1ft | A1: one of awrt 0.84 or awrt $-0.84$; A1ft: apply ft for $x = \pm\alpha$ where $\alpha = \cos^{-1}k$ and $-1 \leq k \leq 1$ |
| $\{\sin x = 0 \Rightarrow\}\ x = 0$ and $-\pi$ | B1 | Both $x=0$ and $-\pi$ or awrt $-3.14$ from $\sin x = 0$; ignore extra solutions in range |

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7. (i) Solve, for $0 \leqslant \theta < 360 ^ { \circ }$, the equation

$$9 \sin \left( \theta + 60 ^ { \circ } \right) = 4$$

giving your answers to 1 decimal place.\\
You must show each step of your working.\\
(ii) Solve, for $- \pi \leqslant x < \pi$, the equation

$$2 \tan x - 3 \sin x = 0$$

giving your answers to 2 decimal places where appropriate. [Solutions based entirely on graphical or numerical methods are not acceptable.]

\hfill \mbox{\textit{Edexcel C2 2014 Q7 [9]}}