Standard +0.3 This is a standard trigonometric equation requiring conversion to quadratic form using tan x = sin x/cos x and the Pythagorean identity, followed by solving a quadratic and applying double angle substitution. While it has multiple steps, each technique is routine for C2 level with no novel insight required, making it slightly easier than average.
10. (a) Given that
$$8 \tan x = - 3 \cos x$$
show that
$$3 \sin ^ { 2 } x - 8 \sin x - 3 = 0$$
(b) Hence solve, for \(0 \leqslant \theta < 360 ^ { \circ }\),
$$8 \tan 2 \theta = - 3 \cos 2 \theta$$
giving your answers to one decimal place.
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\includegraphics[max width=\textwidth, alt={}, center]{53865e15-3838-4551-b507-fe49549b87db-29_124_37_2615_1882}
Use \(\frac{\sin x}{\cos x} = \tan x\) to give \(8\sin x = -3\cos^2 x\)
M1
Or equivalent
Use \(\cos^2 x = 1 - \sin^2 x\) i.e. \(8\sin x = -3(1-\sin^2 x)\)
M1
May also be seen as \(8\tan x = -3\cos x \Rightarrow 8\tan x = -3\sqrt{1-\sin^2 x}\)
So \(8\sin x = -3 + 3\sin^2 x\) and \(3\sin^2 x - 8\sin x - 3 = 0\)
A1*
Proceeds to given answer with no errors. Do not tolerate bracketing or notation errors
Part (b):
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
Solves three term quadratic \(3\sin^2 x - 8\sin x - 3 = 0\)
M1
Solving quadratic by usual methods
So \((\sin x) = -\frac{1}{3}\) (or 3)
A1
Condone \(-\frac{1}{3}\) appearing as awrt \(-0.333\); condone \(x/a/\theta = -\frac{1}{3}\), \(\sin x = -\frac{1}{3}\), \(\sin 2x = -\frac{1}{3}\)
\((2\theta) = -19.47\) or \(199.47\) or \(340.53\)
dM1
Uses inverse sine to obtain answer for \(2\theta\); stipulation \(\lvert k \rvert < 1\); accept rounding to 1dp for \(2\theta\)
\(\theta = 99.7, 170.3, 279.7\) or \(350.3\)
A1, A1
First A1: two correct awrt 1dp; second A1: all four correct awrt 1dp
## Question 10:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\frac{\sin x}{\cos x} = \tan x$ to give $8\sin x = -3\cos^2 x$ | M1 | Or equivalent |
| Use $\cos^2 x = 1 - \sin^2 x$ i.e. $8\sin x = -3(1-\sin^2 x)$ | M1 | May also be seen as $8\tan x = -3\cos x \Rightarrow 8\tan x = -3\sqrt{1-\sin^2 x}$ |
| So $8\sin x = -3 + 3\sin^2 x$ and $3\sin^2 x - 8\sin x - 3 = 0$ | A1* | Proceeds to given answer with no errors. Do not tolerate bracketing or notation errors |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Solves three term quadratic $3\sin^2 x - 8\sin x - 3 = 0$ | M1 | Solving quadratic by usual methods |
| So $(\sin x) = -\frac{1}{3}$ (or 3) | A1 | Condone $-\frac{1}{3}$ appearing as awrt $-0.333$; condone $x/a/\theta = -\frac{1}{3}$, $\sin x = -\frac{1}{3}$, $\sin 2x = -\frac{1}{3}$ |
| $(2\theta) = -19.47$ or $199.47$ or $340.53$ | dM1 | Uses inverse sine to obtain answer for $2\theta$; stipulation $\lvert k \rvert < 1$; accept rounding to 1dp for $2\theta$ |
| $\theta = 99.7, 170.3, 279.7$ or $350.3$ | A1, A1 | First A1: two correct awrt 1dp; second A1: all four correct awrt 1dp |
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10. (a) Given that
$$8 \tan x = - 3 \cos x$$
show that
$$3 \sin ^ { 2 } x - 8 \sin x - 3 = 0$$
(b) Hence solve, for $0 \leqslant \theta < 360 ^ { \circ }$,
$$8 \tan 2 \theta = - 3 \cos 2 \theta$$
giving your answers to one decimal place.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\includegraphics[max width=\textwidth, alt={}, center]{53865e15-3838-4551-b507-fe49549b87db-29_124_37_2615_1882}\\
\hfill \mbox{\textit{Edexcel C12 2016 Q10 [8]}}