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Example 20 - Chapter 3 Class 11 Trigonometric Functions (Term 2)
Last updated at March 22, 2023 by Teachoo
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Chapter 3 Class 11 Trigonometric Functions
Serial order wise
- Ex 3.1
- Ex 3.2
- Ex 3.3
- Ex 3.4
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Examples
- Miscellaneous
Transcript
Example 20 Find the solution of sin x = – √3/2 Let sin x = sin y also sin x = (−√3)/2 From (1) and (2) sin y = (−√3)/2 sin y = sin 4/3 π ⇒ y = 4/3 π Step 2 Since sin x = sin y General Solution is x = n π + (–1)n y where n ∈ Z Putting y = 4𝜋/3 x = n π + (–1)n 4/3 π where n ∈ Z
