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Last updated at Feb. 13, 2020 by Teachoo

Transcript

Example 19 Find the principal solutions of the equation tan x = – 1/√3 . tan x = – 1/√3 We know that tan 30° = 1/√3 Since tan x is negative So, x will lie in llnd and lVth Quadrant Value in llnd Quadrant = 180 – 30° Value in lVth Quadrant = 360 – 30° So Principal Solutions are x = 150° x = 150/180 π x = 𝟓/𝟔 π x = 330° x = 330/180 π x = 𝟏𝟏/𝟔 π

Chapter 3 Class 11 Trigonometric Functions

Concept wise

- Radian measure - Conversion
- Arc length
- Finding Value of trignometric functions, given other functions
- Finding Value of trignometric functions, given angle
- (x + y) formula
- 2x 3x formula - Proving
- 2x 3x formula - Finding value
- cos x + cos y formula
- 2 sin x sin y formula
- Finding Principal solutions
- Finding General Solutions
- Sine and Cosine Formula

About the Author

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.