![Ex 3.4, 4 - Chapter 3 Class 11 Trigonometric Functions - Part 2](https://d1avenlh0i1xmr.cloudfront.net/cb719d8a-441a-4bbf-bd88-f045d9ee4a95/slide10.jpg)
![Ex 3.4, 4 - Chapter 3 Class 11 Trigonometric Functions - Part 3](https://d1avenlh0i1xmr.cloudfront.net/5537eb6d-3ac8-4ce0-8b12-cfef0526d144/slide11.jpg)
![Ex 3.4, 4 - Chapter 3 Class 11 Trigonometric Functions - Part 4](https://d1avenlh0i1xmr.cloudfront.net/68dded92-8f2d-4a79-9b36-4a270287939f/slide12.jpg)
Principal and General Solutions
Principal and General Solutions
Last updated at April 16, 2024 by Teachoo
Question 4 Find the principal and general solution of cosec x = β2 Given cosec x = β2 1/sinβ‘π₯ = β2 sin x = (β1)/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° β 30Β° = 330Β° So, Principal Solutions are Finding general solution Let sin x = sin y sin x = β 1/2 From (1) and (2) sin y = β 1/2 sin y = sin 7π/6 y = 7π/6 (Calculated sππ 7π/6 = (β1)/2 while finding principal solutions) Since sin x = sin y General Solution is x = nΟ + (β1)n y where n β Z Putting y = 7π/6 x = nΟ + (β1)n ππ /π Where n β Z