Solve the following equations.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2015). The mark or mark type is shown in brackets.

1. [M1] (i) tan 2x = -5/4 For obtaining and using tan 2x = ±5/4 or ±4/5 resulting in 2x =
2. [A1] (i) (2x=128.7, 308.7)
3. [A1FT] (i) x = 64.3 awrt ; 154.3 awrt tanx = ... gets M0. their 64.3° + 90°
4. [B1] (ii) cosec²y + 3cosecy – 4 = 0 or 4sin²y-3siny-1=0 In any form as a three term quadratic.
5. [M1] (ii) (cosecy + 4)(cosecy-1)=0 or (4sin y + 1)(sin y − 1) = 0 Solve three term quadratic in cosec y or sin y
6. [A1A1A1] (ii) sin y = -1/4 or sin y = 1 ; y = 194.5, 345.5, 90 Answers must be obtained from the correct quadratic
7. [B1] (iii) z + π/4 = π - π/3 or z + π/4 = π + π/3 Accept 2.09, 2.10, π−1.05, π−1.04 on RHS. Could be implied by final answer
8. [B1] (iii) Accept 4.19, 4.18, π +1.05, π +1.04 on RHS. Could be implied by final answer
9. [B1B1] (iii) z = 5π/12 ; 13π/12 Answers must be correct multiples of π.