Standard +0.3 This is a standard trigonometric equation requiring the Pythagorean identity to convert to a single variable, then solving a quadratic in sin x. The restricted domain and rounding requirement are routine. Slightly easier than average due to straightforward algebraic manipulation and no unusual angles.
Find all the solutions of
$$9 \sin^2 x - 6 \sin x + \cos^2 x = 0$$
where \(0° \leq x \leq 180°\)
Give your solutions to the nearest degree.
Fully justify your answer.
[4 marks]
Question 4:
4 | Uses substitution cos2x = 1 – sin2 x
in any form | 1.2 | B1 | 9sin2x – 6sinx + (1 – sin2x) = 0
8sin2x – 6sinx + 1 = 0
(4sinx – 1)(2sinx – 1)
sin x = ¼
sin x = ½
14°, 30°, 150°, 166°
Solves ‘their’ quadratic to obtain
two values for sin x | 1.1a | M1
Finds two correct solutions for x | 1.1b | A1
Finds all four solutions for x and no
extras
(condone 14.5, 165.5 AWRT) | 1.1b | A1
Total | 4
Q | Marking Instructions | AO | Marks | Typical Solution
Find all the solutions of
$$9 \sin^2 x - 6 \sin x + \cos^2 x = 0$$
where $0° \leq x \leq 180°$
Give your solutions to the nearest degree.
Fully justify your answer.
[4 marks]
\hfill \mbox{\textit{AQA AS Paper 2 2020 Q4 [4]}}