Edexcel C12 2019 June — Question 12 8 marks

Exam BoardEdexcel
ModuleC12 (Core Mathematics 1 & 2)
Year2019
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicQuadratic trigonometric equations
TypeShow then solve: sin²/cos² substitution
DifficultyStandard +0.3 This is a straightforward C2-level trigonometric equation requiring standard techniques: cross-multiply, use sin²x + cos²x = 1 to convert to a quadratic in cos x, then solve the quadratic and find angles. The algebraic manipulation is routine and the question explicitly guides students through the process in two parts. Slightly easier than average due to the scaffolding provided in part (a).
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals

12. (a) Show that $$\frac { 2 + \cos x } { 3 + \sin ^ { 2 } x } = \frac { 4 } { 7 }$$ may be expressed in the form $$a \cos ^ { 2 } x + b \cos x + c = 0$$ where \(a , b\) and \(c\) are constants to be found.
(b) Hence solve, for \(0 \leqslant x < 2 \pi\), the equation $$\frac { 2 + \cos x } { 3 + \sin ^ { 2 } x } = \frac { 4 } { 7 }$$ giving your answers in radians to 2 decimal places.
(Solutions based entirely on graphical or numerical methods are not acceptable.) \includegraphics[max width=\textwidth, alt={}, center]{de511cb3-35c7-4225-b459-a136b6304b78-37_81_65_2640_1886}

Question 12:
Part (a):
AnswerMarks Guidance
WorkingMark Guidance
\(14 + 7\cos x = 4\sin^2 x + 12\)M1 Multiply both sides by \((3 + \sin^2 x)\); condone invisible brackets
\(14 + 7\cos x = 4(1 - \cos^2 x) + 12\)M1 Use \(\sin^2 x = 1 - \cos^2 x\)
\(4\cos^2 x + 7\cos x - 2 = 0\)A1 Or exact equivalent; withhold for missing brackets
Part (b):
AnswerMarks Guidance
WorkingMark Guidance
\((4\cos x - 1)(\cos x + 2) = 0 \Rightarrow \cos x = ...\)M1 Solve quadratic; \(
\(\cos x = \frac{1}{4}\)A1 Ignore \(\cos x = -2\); must come from correct equation
\(x = \cos^{-1}\left(\frac{1}{4}\right) = ...\)dM1 Find \(\arccos(\alpha)\); dependent on previous M1
\(x =\) awrt \(1.32\), awrt \(4.96/4.97\)A1A1 Both required; withhold if additional angles in range. Answers with no working score 0
## Question 12:

### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $14 + 7\cos x = 4\sin^2 x + 12$ | M1 | Multiply both sides by $(3 + \sin^2 x)$; condone invisible brackets |
| $14 + 7\cos x = 4(1 - \cos^2 x) + 12$ | M1 | Use $\sin^2 x = 1 - \cos^2 x$ |
| $4\cos^2 x + 7\cos x - 2 = 0$ | A1 | Or exact equivalent; withhold for missing brackets |

### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $(4\cos x - 1)(\cos x + 2) = 0 \Rightarrow \cos x = ...$ | M1 | Solve quadratic; $|\alpha| \leq 1$ required |
| $\cos x = \frac{1}{4}$ | A1 | Ignore $\cos x = -2$; must come from correct equation |
| $x = \cos^{-1}\left(\frac{1}{4}\right) = ...$ | dM1 | Find $\arccos(\alpha)$; dependent on previous M1 |
| $x =$ awrt $1.32$, awrt $4.96/4.97$ | A1A1 | Both required; withhold if additional angles in range. Answers with no working score 0 |
12. (a) Show that

$$\frac { 2 + \cos x } { 3 + \sin ^ { 2 } x } = \frac { 4 } { 7 }$$

may be expressed in the form

$$a \cos ^ { 2 } x + b \cos x + c = 0$$

where $a , b$ and $c$ are constants to be found.\\
(b) Hence solve, for $0 \leqslant x < 2 \pi$, the equation

$$\frac { 2 + \cos x } { 3 + \sin ^ { 2 } x } = \frac { 4 } { 7 }$$

giving your answers in radians to 2 decimal places.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)\\

\includegraphics[max width=\textwidth, alt={}, center]{de511cb3-35c7-4225-b459-a136b6304b78-37_81_65_2640_1886}\\

\hfill \mbox{\textit{Edexcel C12 2019 Q12 [8]}}