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Last updated at Feb. 12, 2020 by Teachoo

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Ex 3.4, 4 Find the principal and general solution of cosec x = β2 Given cosec x = β2 1/sinβ‘π₯ = β2 sin x = (β1)/2 We know that sin 30Β° = 1/2 Since sin x is negative, x will be in lllrd & lVth Quadrant Value in IIIrd Quadrant = 180Β° + 30Β° = 210Β° Value in IVth Quadrant = 360Β° β 30Β° = 330Β° So, Principal Solutions are Finding general solution Let sin x = sin y sin x = β 1/2 From (1) and (2) sin y = β 1/2 sin y = sin 7π/6 y = 7π/6 (Calculated sππ 7π/6 = (β1)/2 while finding principal solutions) Since sin x = sin y General Solution is x = nΟ + (β1)n y where n β Z Putting y = 7π/6 x = nΟ + (β1)n ππ /π Where n β Z

Ex 3.4

Ex 3.4, 1
Not in Syllabus - CBSE Exams 2021

Ex 3.4, 2 Not in Syllabus - CBSE Exams 2021

Ex 3.4, 3 Important Not in Syllabus - CBSE Exams 2021

Ex 3.4, 4 Important Not in Syllabus - CBSE Exams 2021 You are here

Ex 3.4, 5 Important Not in Syllabus - CBSE Exams 2021

Ex 3.4, 6 Important Not in Syllabus - CBSE Exams 2021

Ex 3.4, 7 Important Not in Syllabus - CBSE Exams 2021

Ex 3.4, 8 Important Not in Syllabus - CBSE Exams 2021

Ex 3.4, 9 Important Not in Syllabus - CBSE Exams 2021

Chapter 3 Class 11 Trigonometric Functions

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.