# Plus One Math's Solution Ex 3.4 Chapter 3 Trigonometric Functions

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The solutions for the
Trigonometric Functions are not readily available. While many schools may have a ready-made solution for this set in their school textbook, some might not. This is where our solution will be useful. In this article, you will find detailed solutions provided by us for the above set.Trigonometric Functions (Key Concept Reference) describes some basic and advanced uses of trigonometric functions, including identities, graph transformations, inverse functions, solutions of triangles, and polar coordinates.

Ncert Plus one Maths chapter-wise textbook solution for chapter 3 Trigonometric Functions Exercise 3.4. It contains detailed solutions for each question which have prepared by expert teachers to make each answer easily understand the students. they are well arranged solutions so that students would be able to understand easily.

 Board SCERT, Kerala Text Book NCERT Based Class Plus One Subject Math's Textbook Solution Chapter Chapter 3 Exercise Ex 3.4 Chapter Name Trigonometric Functions Category Plus One Kerala

## Kerala Syllabus Plus One Math's Textbook Solution Chapter  3 Trigonometric Functions Exercises 3.4

### Chapter  3 Trigonometric Functions Textbook Solution

Kerala plus One maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.

### Chapter  3 Trigonometric Functions Exercise   3.4

Question 1:

Find the principal and general solutions of the equation   Therefore, the principal solutions are x = and . Therefore, the general solution is Find the principal and general solutions of the equation sec x=2

cos x=1/secx=1/2

x=Π/3

cos(2Π-x)=cosx

cos(2Π-Π/3)=cos(5Π/3)=1/2

principal solutions are Π/3 and 5Π/3

cosx=cosΠ/3

x=2nΠ±Π/3 ,n∈Z, is the general solution

Find the principal and general solutions of the equation  tanx=-1/√3

tan Π/6=1/√3

tan(Π-x)=-tanx=-1/√3

tan(Π-Π/6)=tan(5Π/6)=-1/√3

tan(2Π-x)=-tanx=-1/√3

tan(2Π-Π/6)=tan(11Π/6)=-1/√3

principal solutions are 5Π/6 and 11Π/6

tanx=tan5Π/6

x=nΠ+5Π/6, n∈Z is the general solution

Find the general solution of cosec x = –2

cosec x = –2 Therefore, the principal solutions are x = . Therefore, the general solution is Find the general solution of the equation   Find the general solution of the equation    Find the general solution of the equation   Therefore, the general solution is .

Find the general solution of the equation    Therefore, the general solution is .

Find the general solution of the equation    Therefore, the general solution is #### Chapter 3 Trigonometric Functions EX 3.4 Solution- Preview

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