O-LevelAdditional MathematicsTrigonometry4037/12 · May/June 2011 · Q1111 Marks
Solve
- (i)2cot² x – 5cosec x−1=0 for 0° < x < 180°,[4]
- (ii)5cos 2y - 4sin 2y = 0 for 0° < y < 180°,[4]
- (iii)cos (z + π/6) = -½ for 0 < z < 2π radians.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.
- (i) 2(cosec²x – 1) – 5cosec x − 1 = 0; M1 for use of correct identity or to get in terms of sin x
- (2cosec x + 1)(cosec x – 3) = 0; DM1 for attempt to solve
- leading to sin x = ⅓; A1
- x = 19.5°, 160.5°; √A1 180° – their x
- (ii) tan 2y = 5/4; M1 for attempt to get in terms of tan
- 2y = 51.34°, 231.34°; M1 for dealing correctly with double angle
- y = 25.7°, 115.7°; A1√A1 √90° their y
- (iii) (z + π/6) = (2π/3), (4π/3); M1 for dealing with order correctly and attempt to solve
- z = (2π/3) – (π/6) , z = (4π/3) – (π/6); A1, A1
- allow 1.57, 3.67
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