


Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Principal and General Solutions
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams You are here
Question 5 Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 8 Important Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Principal and General Solutions
Last updated at May 29, 2023 by Teachoo
Question 4 Find the principal and general solution of cosec x = –2 Given cosec x = –2 1/sin𝑥 = −2 sin x = (−1)/2 We know that sin 30° = 1/2 Since sin x is negative, x will be in lllrd & lVth Quadrant Value in IIIrd Quadrant = 180° + 30° = 210° Value in IVth Quadrant = 360° – 30° = 330° 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/6 y = 7𝜋/6 (Calculated s𝑖𝑛 7𝜋/6 = (−1)/2 while finding principal solutions) Since sin x = sin y General Solution is x = nπ + (−1)n y where n ∈ Z Putting y = 7𝜋/6 x = nπ + (−1)n 𝟕𝝅/𝟔 Where n ∈ Z