Ex 3.4, 4 - cosec x = -2, find principal and general solution

Ex 3.4, 4 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Ex 3.4, 4 - Chapter 3 Class 11 Trigonometric Functions - Part 3 Ex 3.4, 4 - Chapter 3 Class 11 Trigonometric Functions - Part 4

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Question 4 Find the principal and general solution of cosec x = –2 Given cosec x = –2 1/sin⁑π‘₯ = βˆ’2 sin x = (βˆ’1)/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° – 30Β° = 330Β° So, Principal Solutions are Finding general solution Let sin x = sin y sin x = – 1/2 From (1) and (2) sin y = – 1/2 sin y = sin 7πœ‹/6 y = 7πœ‹/6 (Calculated s𝑖𝑛 7πœ‹/6 = (βˆ’1)/2 while finding principal solutions) Since sin x = sin y General Solution is x = nΟ€ + (βˆ’1)n y where n ∈ Z Putting y = 7πœ‹/6 x = nΟ€ + (βˆ’1)n πŸ•π…/πŸ” Where n ∈ Z

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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.