A student attempts to solve the equation cos x + sin x tan x = 2sin x − 1 in the range 0 ≤ x ≤ 2π.
The student's attempt is as follows:
cos x + sin x tan x = 2 sin x − 1
So cos x − sin x + sin x tan x − sin x = −1 (I)
So (sin x − cos x)(tan x − 1) = −1 (II)
So sin x − cos x = −1 or tan x − 1 = −1 (III)
So (sin x − cos x)2 = 1 or tan x = 0 (IV)
So 2sin x cos x = 0 or tan x = 0 (V)
So x = 0, π2, π, 3π2, 2π (VI)
Which of the following best describes this attempt?
- AIt is completely correct.
- BIt is incorrect, and the first error occurs on line (I)
- CIt is incorrect, and the first error occurs on line (II)
- DIt is incorrect, and the first error occurs on line (III)
- EIt is incorrect, and the first error is that extra solutions were introduced on line (IV)
- FIt is incorrect, and the first error is that extra solutions were introduced on line (V)
- GIt is incorrect, and the first error is not eliminating the values where tan x is undefined on line (VI)
Show the answer and worked solution
answer · D
- AIt is completely correct.
- BIt is incorrect, and the first error occurs on line (I)
- CIt is incorrect, and the first error occurs on line (II)
- DIt is incorrect, and the first error occurs on line (III)
- EIt is incorrect, and the first error is that extra solutions were introduced on line (IV)
- FIt is incorrect, and the first error is that extra solutions were introduced on line (V)
- GIt is incorrect, and the first error is not eliminating the values where tan x is undefined on line (VI)
Check the first two lines by expanding rather than by eye. Line (I) is just 2sin x taken across as −sin x − sin x, so it is fine. For line (II), (sin x − cos x)(tan x − 1) = sin xtan x − sin x − cos xtan x + cos x, and cos x tan x = sin x, giving exactly the left-hand side of (I). Line (III) is where it breaks: from a product equal to −1 you cannot conclude that one factor is −1, because two numbers can multiply to −1 without either being −1 (the split-into-cases move only works when the product is 0).