| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Session | Specimen |
| Marks | 4 |
| Topic | Standard trigonometric equations |
| Type | Convert to quadratic in sin/cos |
| Difficulty | Moderate -0.3 This is a standard trigonometric equation requiring routine algebraic manipulation (converting cosec to sin, using cos²x = 1 - sin²x) followed by solving a straightforward quadratic. The 'show that' part guides students through the conversion, making it slightly easier than average for A-level. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
**(i)** Forming a quadratic in $\sin x$ using identities M1
$\cos^2 x = 1 - \sin^2 x$ and $\text{cosec}\, x = \frac{1}{\sin x}$
$4 - \frac{4}{\text{cosec}\, x} = 3(1 - \sin^2 x)$
$4 - 4\sin x = 3(1 - \sin^2 x)$
$3\sin^2 x - 4\sin x + 1 = 0$ All correctly deduced A1 **[2]**
**(ii)** Attempt solution of a quadratic in $\sin x$ M1
$\sin x = \frac{1}{3}$ $\quad$ $\sin x = 1$ Both solutions A1
$x = 0.340$ $\quad$ $x = \frac{\pi}{2}$ *OR* $1.57$ A1
$x = 2.80$ (f.t. on their primary value) A1$\checkmark$ **[4]**
SR Answers in degrees (90, 19.5, 160.5) lose last A15 (i) Show that the equation $4 - \frac { 4 } { \operatorname { cosec } x } = 3 \cos ^ { 2 } x$ can be expressed in the form
$$3 \sin ^ { 2 } x - 4 \sin x + 1 = 0$$
(ii) Hence find all values of $x$ for which $0 < x < \pi$ that satisfy the equation
$$4 - \frac { 4 } { \operatorname { cosec } x } = 3 \cos ^ { 2 } x$$
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 Q5 [4]}}