TTMUA Lab
TMUA 2017 · Paper 1 · Question 8 of 20

TMUA 2017 Paper 1 Question 8

Trigonometry — Trigonometric inequality · sign of a product. Try it first; the answer and a full worked solution are below.

TMUA 2017 · Paper 1TrigonometryTrigonometric inequality · sign of a product4 options
Find the complete set of values of x, with 0 ≤ x ≤ π, for which (1 − 2sin x)cos x ≥ 0
  1. A0 ≤ x ≤ π6,   π2 ≤ x ≤ 5π6
  2. B0 ≤ x ≤ π6,   5π6 ≤ x ≤ π
  3. Cπ6 ≤ x ≤ π2,   5π6 ≤ x ≤ π
  4. Dπ6 ≤ x ≤ 5π6
Show the answer and worked solution
answer · A
  1. A0 ≤ x ≤ π6,   π2 ≤ x ≤ 5π6
  2. B0 ≤ x ≤ π6,   5π6 ≤ x ≤ π
  3. Cπ6 ≤ x ≤ π2,   5π6 ≤ x ≤ π
  4. Dπ6 ≤ x ≤ 5π6
A product is non-negative when both factors have the same sign, so treat the two factors separately on [0, π]. Here 1 − 2sin x ≥ 0 means sin x ≤ 12, which holds on [0,  π6] and on [5π6,  π]; and cos x ≥ 0 holds on [0,  π2]. Both non-negative gives 0 ≤ x ≤ π6. Both non-positive needs π6 ≤ x ≤ 5π6 together with π2 ≤ x ≤ π, that is π2 ≤ x ≤ 5π6.