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

Transcript

Ex 3.4, 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
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.