Lemmara logoLemmaraWorked solution

0580/42 Feb/Mar 2023, Question 12: Trigonometric Graphs and Equations

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 4 marks

اقرأ هذا الحل مشروحاً بالعربي ←

The question, in short

Sketch y = tan x over 0° to 360°, then solve tan x = 1/√3 within the same interval.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 2 marks

Draw the tangent curve for 0° ≤ x ≤ 360°, showing behaviour at 90°, 180°, 270° and 360°.

  1. 1Start at (0, 0) and rise steeply towards the asymptote x = 90°
  2. 2Middle branch comes up from below at 90°, passes through (180, 0), rises towards x = 270°
  3. 3Last branch comes up from below at 270° and reaches (360, 0)

AnswerCorrect sketch of y = tan x: passing through (0, 0), (180, 0) and (360, 0), increasing branches with asymptotes at x = 90° and x = 270°

Part (b) · 2 marks

Find all values of x between 0° and 360° satisfying tan x = 1/√3.

  1. 1tan 30° = 1/√3, so x = 30°
  2. 2tan repeats every 180°
  3. 3x = 30 + 180 = 210°
  4. 4x = 30°, 210°

Answer30 and 210

Stuck on a step? Read the free lesson: Trigonometric Graphs and Equations

Practise more trigonometric graphs and equations questions

Real 0580 past-paper questions, marked on your working like the exam, with a worked solution for every part.

Practise free on Lemmara

More past-paper questions on trigonometric graphs and equations

Also searched as: 0580/42/F/M/23 · 0580_m23_qp_42 · 0580 m23 42 · question 12.
Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education. The worked solutions are Lemmara’s own.