Date | May 2021 | Marks available | 5 | Reference code | 21M.1.SL.TZ2.3 |
Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 2 |
Command term | Solve and Hence | Question number | 3 | Adapted from | N/A |
Question
Show that the equation 2 cos2 x+5 sin x=42cos2x+5sinx=4 may be written in the form 2 sin2 x-5 sin x+2=0.
Hence, solve the equation 2 cos2 x+5 sin x=4, 0≤x≤2π.
Markscheme
METHOD 1
correct substitution of cos2 x=1- sin2 x A1
2(1-sin2 x)+5 sin x=4
2 sin2 x-5 sin x+2=0 AG
METHOD 2
correct substitution using double-angle identities A1
(2 cos2 x-1)+5 sin x=3
1-2 sin2 x-5 sin x=3
2 sin2 x-5 sin x+2=0 AG
[1 mark]
EITHER
attempting to factorise M1
(2 sin x−1)(sin x−2) A1
OR
attempting to use the quadratic formula M1
sin x=5±√52-4×2×24(=5±34) A1
THEN
sin x=12 (A1)
x=π6, 5π6 A1A1
[5 marks]