A-LevelMathematicsTrigonometry9709/12 · May/June 2018 · Q108 Marks

Solve the equation 2 cos x + 3 sin x = 0, for 0° ≤ x ≤ 360°.

(i)Solve the equation 2 cos x + 3 sin x = 0, for 0° ≤ x ≤ 360°.[3]
(ii)Sketch, on the same diagram, the graphs of y = 2 cos x and y = −3 sin x for 0° ≤ x ≤ 360°.[3]
(iii)Use your answers to parts (i) and (ii) to find the set of values of x for 0° ≤ x ≤ 360° for which 2 cos x + 3 sin x > 0.[2]

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

[M1] (i) To solve 2 cos x + 3 sin x = 0, rearrange to 3 sin x = -2 cos x. Divide by cos x (assuming cos x ≠ 0): tan x = -2/3. Use of tan=sin/cos to get tan =, or other valid method to find sin or cos. M0 for tanx = +/- 3/2.
[A1 A1FT] (i) The principal value for tan x = -2/3 is arctan(-2/3) ≈ -33.69°. Since tan x is negative, x lies in the 2nd and 4th quadrants. So, x = 180° - 33.69° = 146.3° (to 1 d.p.) and x = 360° - 33.69° = 326.3° (to 1 d.p.). FT for 180 added to an incorrect first answer in the given range. The second A1 is withheld if any further values in the range 0°≤ x ≤ 360° are given. Answers in radians score A0, A0.
[B1] (ii) Sketch of y = 2 cos x. One complete cycle; start and finish at top of curve at roughly the same positive y value and go below the x axis by roughly the same distance. (Can be a poor curve but not straight lines.)
[B1] (ii) Sketch of y = -3 sin x. One complete cycle; start and finish on the x axis, must be inverted and go below and then above the x axis by roughly the same distance. (Can be a poor curve but not straight lines.)
[B1] (ii) Fully correct answer including the sine curve with clearly larger amplitude than cosine curve. Must now be reasonable curves. Note: Separate diagrams can score 2/3.
[B1FT B1FT] (iii) The inequality 2 cos x + 3 sin x > 0 is equivalent to 3 sin x > -2 cos x, or dividing by cos x (and considering the sign of cos x), tan x > -2/3 or tan x < -2/3. Graphically, this means finding where the graph of y = -3 sin x is above the graph of y = 2 cos x. From part (i), the intersection points are x = 146.3° and x = 326.3°. By observing the graph (or testing values), y = -3 sin x is above y = 2 cos x when 0° ≤ x < 146.3° and 326.3° < x ≤ 360°. Does not need to include 0°, 360°. √ from their answers in (i) Allow combined statement as long as correct inequalities if taken separately. SC For two correct values including ft but with < and > B1.