Question 2 - Examples - Chapter 3 Class 11 Trigonometric Functions
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 19 Find the principal solutions of the equation tan x = – 1/√3 . tan x = – 1/√3 We know that tan 30° = 1/√3 Since tan x is negative So, x will lie in llnd and lVth Quadrant Value in llnd Quadrant = 180 – 30° Value in lVth Quadrant = 360 – 30° So Principal Solutions are x = 150° x = 150/180 π x = 𝟓/𝟔 π x = 330° x = 330/180 π x = 𝟏𝟏/𝟔 π
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo