Chapter 3 Class 11 Trigonometric Functions
Ex 3.1, 2 (i)
Ex 3.2, 7 Important
Ex 3.2, 8
Ex 3.2, 9 Important
Ex 3.3, 4
Ex 3.3, 5 (i) Important
Ex 3.3, 8 Important
Ex 3.3, 11 Important
Ex 3.3, 18 Important
Ex 3.3, 23 Important
Ex 3.3, 21 Important
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams You are here
Question 8 Important Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Example 20 Important
Example 21 Important
Misc 4 Important
Misc 7 Important
Chapter 3 Class 11 Trigonometric Functions
Last updated at April 16, 2024 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