Principal and General Solutions

Chapter 3 Class 11 Trigonometric Functions
Serial order wise

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### Transcript

Question 3 Find the principal and general solutions of the equation cot x = –√3 Given cot x = −√3 tan x = 1/cot⁡𝑥 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° = 150° Value in lVth Quadrant = 360° – 30° = 330° So, Principal Solution are x = 150° x = 150 × 𝜋/180 x = 𝟓𝝅/𝟔 x = 330° x = 330 × 𝜋/180 x = 𝟏𝟏𝝅/𝟔 Finding general solution Let tan x = tan y tan x = (−1)/√3 Form (1) and (2) tan y = (−1)/√3 tan y = tan 5𝜋/6 y = 5𝜋/6 (Calculated tan 5𝜋/6 = (−1)/√3 while finding principal solutions) Since tan x = tan y General Solution is x = nπ + y where n ∈ Z Putting x = 5𝜋/6 x = nπ + 𝟓𝝅/𝟔 where n ∈ Z

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.