A-LevelMathematicsTrigonometry9709/12 · Feb/Mar 2020 · Q119 Marks

Solve the equation 3 tan²x – 5 tan x − 2 = 0 for 0° ≤ x ≤ 180°.

(a)Solve the equation 3 tan²x – 5 tan x − 2 = 0 for 0° ≤ x ≤ 180°.[4]
(b)Find the set of values of k for which the equation 3 tan²x – 5 tanx + k = 0 has no solutions.[2]
(c)For the equation 3 tan²x – 5 tan x + k = 0, state the value of k for which there are three solutions in the interval 0° ≤ x ≤ 180°, and find these solutions.[3]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2020). The mark or mark type is shown in brackets.

[M1] (a) (tan x - 2)(3 tan x + 1) (= 0). or formula or completing square Allow reversal of signs in the factors. Must see a method
[A1] (a) tan x = 2 or -1/3
[B1FT] (a) x = 63.4°(only value in range)
[B1FT] (a) x = 161.6°(only value in range)
[M1] (b) Apply b² - 4ac < 0 SOI. Expect 25-4(3)(k) <0, tan x must not be in coefficients
[A1] (b) k > 25/12 Allow b² – 4ac = 0 leading to correct k > 25/12 for M1A1
[M1] (c) k = 0 SOI
[A1] (c) tan x = 0 or 5/3
[A1] (c) x = 0° or 180° or 59.0° All three required

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