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

  1. Chapter 3 Class 11 Trigonometric Functions
  2. Serial order wise
Ask Download

Transcript

Ex 3.4, 4 Find the principal and general solution of cosec x = –2 Given cosec x = –2 β‡’ 1/sin⁑π‘₯ = -2 sin x = (βˆ’1)/2 We know that sin 30Β° = 1/2 But we need negative value of sin x sin x is negative in lllrd & lVth Quadrant Value in IIIrd Quadrant = 180Β° + 30Β° = 210Β° Value in IVth Quadrant = 360Β° – 30Β° = 330Β° So, Principle Solution are x = 210Β° and x = 330Β° x = 210 Γ— πœ‹/180 and x = 330 Γ— πœ‹/180 x = 7πœ‹/6 and x = 11πœ‹/6 To find 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 Since sin x = sin y General Solution is x = nΟ€ + ( -1)n y where n ∈ Z Put y = 7πœ‹/6 Hence, x = nΟ€ +(-1)n 7πœ‹/6 Where n ∈ Z

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
Jail