NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions (Ex 3.4) Exercise 3.4

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions (Ex 3.4) Exercise 3.4

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric functions are provided by Vedantu for the best preparation for your exams. Vedantu provides detailed and reliable solutions for Class 11 Maths NCERT solutions trigonometry exercise 3.4. Through Vedantu's online web portal, students can have a proper understanding of the concepts for each and every topic of Class 11 NCERT Trigonometry. Additionally, students can also download the NCERT Solutions Class 11 Maths Chapter 3 free pdf study material from Vedantu's website and refer offline.

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Exercise 3.4

1. Find the principal and general solutions of the $\text{tan x=}\sqrt{\text{3}}$.

Ans: Here given that,

$\text{tan x=}\sqrt{\text{3}}$

We know that $\text{tan}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=}\sqrt{\text{3}}$

and $\text{tan}\left( \frac{\text{4 }\!\!\pi\!\!\text{ }}{\text{3}} \right)\text{=tan}\left( \text{ }\!\!\pi\!\!\text{ +}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)\text{=tan}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=}\sqrt{\text{3}}$

Therefore, the principal solutions are$\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ and $\frac{\text{4 }\!\!\pi\!\!\text{ }}{\text{3}}$ .

Now, $\text{tan x=tan}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

Which implies,

$\text{x=n }\!\!\pi\!\!\text{ +}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ , where $\text{n}\in \text{Z}$

Therefore, the general solution is $\text{x=n }\!\!\pi\!\!\text{ +}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$, where $\text{n}\in \text{Z}$.

2. Find the principal and general solutions of the equation $\text{secx=2}$

Ans: Here it is given that,

$\text{sec x=2}$

Now we know that

$\text{sec}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{=2}$ and

$\text{sec}\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}\text{=sec}\left( \text{2 }\!\!\pi\!\!\text{ -}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

$\text{=sec}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

$\text{=2}$

Therefore, the principal solutions are$\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ and $\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{3}}$.

Now, $\text{sec x=sec}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

and we know ,

$\sec x=\frac{1}{\cos x}$

Therefore , we have,

$\text{cos x=cos}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

Which implies,

$\text{x=2n }\!\!\pi\!\!\text{ }\!\!\pm\!\!\text{ }\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ , where $\text{n}\in \text{Z}$ .

Therefore, the general solution is $\text{x=2n }\!\!\pi\!\!\text{ }\!\!\pm\!\!\text{ }\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ , where $n\in Z$.

3. Find the principal and general solutions of the equation $\text{cot x=-}\sqrt{\text{3}}$

Ans: Here it is given that,

$\text{cot x=-}\sqrt{\text{3}}$

Now we know that $\text{cot}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{=}\sqrt{\text{3}}$

And

$\text{cot}\left( \text{ }\!\!\pi\!\!\text{ -}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cot}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

$\text{=-}\sqrt{\text{3}}$

and  $\text{cot}\left( \text{2 }\!\!\pi\!\!\text{ -}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cot}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

$\text{=-}\sqrt{\text{3}}$

Therefore we have,

$\text{cot}\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-}\sqrt{\text{3}}$

and $\text{cot}\frac{\text{11 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-}\sqrt{\text{3}}$

Therefore, the principal solutions are $\text{x=}\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$ and $\frac{\text{11 }\!\!\pi\!\!\text{ }}{\text{6}}$.

Now, $\text{cot x=cot}\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$

And we know $\text{cot x=}\frac{\text{1}}{\text{tan x}}$

Therefore we have,

$\text{tan x=tan}\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$

Which implies,

$\text{x=n }\!\!\pi\!\!\text{ +}\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$ , where $\text{n}\in \text{Z}$

Therefore, the general solution is $\text{x=n }\!\!\pi\!\!\text{ +}\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$ , where $\text{n}\in \text{Z}$.

4. Find the general solution of $\text{cosec x=-2}$

Ans: Here it is given that,

$\text{cosec x=-2}$

Now we know that

$\text{cosec}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{=2}$

and

$\text{cosec}\left( \text{ }\!\!\pi\!\!\text{ +}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cosec}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{a}$

$\text{=-2}$

and $\text{cosec}\left( \text{2 }\!\!\pi\!\!\text{ -}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=-cosec}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

$\text{=-2}$

therefore we have,

$\text{cosec}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-2}$and $\text{cosec}\frac{\text{11 }\!\!\pi\!\!\text{ }}{\text{6}}\text{=-2}$

Hence , the principal solutions are$\text{x=}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\,$ and $\text{ }\frac{11\pi }{6}$.

Now, $\text{cosec x=cosec}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

And we know, $\text{cosec x=}\frac{\text{1}}{\text{sin x}}$

Therefore , we have,

$\text{sin x=sin}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

Which implies,

$\text{x=n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

,where $\text{n}\in \text{Z}$.

Therefore, the general solution is $\text{x=n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{ }$ ,where $\text{n}\in \text{Z}$.

5. Find the general solution of the equation $\text{cos 4x=cos 2x}$

Ans: Here it is given that, $\text{cos 4x=cos 2x}$

Which implies,

$\text{cos 4x-cos 2x=0}$

Now we know that, $\text{cos A-cos B=-2sin}\left( \frac{\text{A+B}}{\text{2}} \right)\text{sin}\left( \frac{\text{A-B}}{\text{2}} \right)$

Therefore we have,

$\text{-2sin}\left( \frac{\text{4x+2x}}{\text{2}} \right)\text{sin}\left( \frac{\text{4x-2x}}{\text{2}} \right)\text{=0}$

$\text{sin 3x sin x=0}$

Hence we have, $\text{sin 3x=0}\,\,$

Or, $\text{ sin x=0}$

Therefore, $\text{3x=n }\!\!\pi\!\!\text{ }$

Or $\text{x=n }\!\!\pi\!\!\text{ }$    ,where $\text{ n}\in \text{Z}$

therefore, $\text{x=}\frac{\text{n }\!\!\pi\!\!\text{ }}{\text{3}}$

Or $\text{x=n }\!\!\pi\!\!\text{ }$  ,where $\text{ n}\in \text{Z}$.

6. Find the general solution of the equation $\text{cos 3x+cos x-cos 2x=0}$.

Ans: Here given that,

$\text{cos 3x+cos x-cos 2x=0}$

Now we know that, $\text{cos A+cos B=2cos}\left( \frac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \frac{\text{A-B}}{\text{2}} \right)$

Therefore  $\text{cos 3x+cos x-cos 2x=0}$ implies

$\text{2cos}\left( \frac{\text{3x+x}}{\text{2}} \right)\text{cos}\left( \frac{\text{3x-x}}{\text{2}} \right)\text{-cos 2x=0}$

$\text{2cos 2x cos x-cos 2x=0}$

$\text{cos 2x}\left( \text{2cos x-1} \right)\text{=0}$

Hence we have,

Either $\text{cos 2x=0}$

Or $\text{cos x=}\frac{\text{1}}{\text{2}}$

Which in turn implies that,

Either $\text{2x=}\left( \text{2n+1} \right)\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\,$

Or,  $\text{cos x=cos}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$    , where  $\text{n}\in \text{Z}$

Therefore,

Either $\text{x=}\left( \text{2n+1} \right)\frac{\text{ }\!\!\pi\!\!\text{ }}{4}\,\,$

Or, $\text{x=2n }\!\!\pi\!\!\text{ }\!\!\pm\!\!\text{ }\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$  ,where  $\text{n}\in \text{Z}$.

7. Find the general solution of the equation $\text{sin 2x+cos x=0}$ .

Ans: Here it is given that,

$\text{sin 2x cos x=0}$

Now we know that, $\text{sin 2x=2sin x cos x}$

Therefore we have,

$\text{2sin x cos x+cos x=0}$

Which implies,

$\text{cos x(2sin x+1)= }\!\!~\!\!\text{ 0}$

Therefore we have,

Either $\text{cos x=0}$

Or, $\text{sin x=-}\frac{\text{1}}{\text{2}}$

Hence we have,

Either  $\text{x= }\!\!~\!\!\text{ (2n+1)}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$   , where $\text{n}\in \text{Z}$ .

Or,  $\text{sin x=-}\frac{\text{1}}{\text{2}}$

$\text{=-sin}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

$\text{=sin}\left( \text{ }\!\!\pi\!\!\text{ -}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

$\text{=sin}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

Which implies

$\text{x=n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$  , where  $\text{n}\in \text{Z}$

Therefore, the general solution is $\left( \text{2n+1} \right)\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\,$ or  $\text{n }\!\!\pi\!\!\text{ +}{{\left( \text{-1} \right)}^{\text{n}}}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\text{,n}\in \text{Z}$.

8. Find the general solution of the equation $\text{se}{{\text{c}}^{\text{2}}}\text{2x=1-tan 2x}$

Ans:

Here given that , $\text{se}{{\text{c}}^{\text{2}}}\text{2x=1-tan 2x}$

Now we know that, $\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}$

Therefore we have,

$\text{se}{{\text{c}}^{\text{2}}}\text{2x=1-tan 2x}$  implies

$\text{1+ta}{{\text{n}}^{\text{2}}}\text{2x=1-tan 2x}$

$\text{ta}{{\text{n}}^{\text{2}}}\text{2x+tan 2x=0}$

$\text{tan 2x(tan 2x+1)= }\!\!~\!\!\text{ 0}$

Hence  either $\text{tan 2x=0}$

Or, $\text{tan 2x=-1}$

Which implies  either  $\text{x=}\frac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}$  , where $\text{n}\in \text{Z}$ ,

Or, $\text{tan 2x=-1}$

$\text{=-tan}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

$\text{=tan}\left( \text{ }\!\!\pi\!\!\text{ -}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)$

$\text{=tan}\frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}$

Which in turn implies that,

$\text{2x=n }\!\!\pi\!\!\text{ +}\frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{,}$ where  $\text{n}\in \text{Z}$

i.e,  $\text{x=}\frac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\text{+}\frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{8}}\text{,}$ where $\text{n}\in \text{Z}$.

Therefore, the general solution is $\frac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\,\,$ or  $\,\,\frac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\text{+}\frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{8}}\text{,n}\in \text{Z}$.

9. Find the general solution of the equation $\text{sin x+sin 3x+sin 5x=0}$

Ans:

Here given that ,$\text{sin x+sin 3x+sin 5x=0}$

Now we know that,  $\text{sin A+sin B=2sin}\left( \frac{\text{A+B}}{\text{2}} \right)\text{cos}\left( \frac{\text{A-B}}{\text{2}} \right)$

Therefore ,

$\text{sin x+sin 3x+sin 5x=0}$

$\left( \text{sin x+sin 3x} \right)\text{+sin 5x=0}$

$\left[ \text{2sin}\left( \frac{\text{x+5x}}{\text{2}} \right)\text{cos}\left( \frac{\text{x-5x}}{\text{2}} \right) \right]\text{+sin 3x=0}\,$

$\text{2sin 3x cos (-2x)+sin 3x= }\!\!~\!\!\text{ 0}$

Simplifying we get,

$\text{2sin 3xcos 2x+sin 3x=0}$

$\text{sin 3x(2cos 2x+1)= }\!\!~\!\!\text{ 0}$

Hence either $\text{sin 3x=0}$

Or, $\text{cos 2x=-}\frac{\text{1}}{\text{2}}$

Which implies  $\text{3x=n }\!\!\pi\!\!\text{ }$ , where $\text{n}\in \text{Z}$

Or,   $\text{cos 2x=-}\frac{\text{1}}{\text{2}}$

$\text{=-cos}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

$\text{=cos}\left( \text{ }\!\!\pi\!\!\text{ -}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

$\text{=cos}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}}$

i.e., either $\text{x=}\frac{\text{n }\!\!\pi\!\!\text{ }}{\text{3}}$   , where $\text{n}\in \text{Z}$

or,  $\text{2x=2n }\!\!\pi\!\!\text{ }\!\!\pm\!\!\text{ }\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}}$   ,where $\text{n}\in \text{Z}$ .

Therefore, the general solution is $\frac{\text{n }\!\!\pi\!\!\text{ }}{\text{3}}\,$ or $\text{n }\!\!\pi\!\!\text{ }\!\!\pm\!\!\text{ }\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{,n}\in \text{Z}$.

Introduction To Trigonometry

In this chapter, we are going to study various trigonometric functions like sin, cos, tan, sec, cosec and other different concepts used in Trigonometry, their relation with each other, measures of angle in different ways, principle solution, general solution, theorems related to real numbers x and y, their application and conversions and many more.

• Trigonometry is a branch of mathematics that gives the relation between side length and angle.

• Trigonometry also means a measure of the angle.

• Trigonometry, though it may sound boring, has a lot of applications in the modern world.

Some of them are listed below:

• Architecture: the designing of structures, require a lot of Trigonometric solutions

• Digital imaging: Trigonometric functions help in the detection of tumors, scanning

• Oceanography

• Satellite systems: the navigation study is done using Trigonometry

• Cartography which is basically the creation of maps uses these functions

• Games: the development in the world of games owes its credits to Trigonometry.

Terms Used For Trigonometric Functions:

• Trigonometry: measuring the sides of a triangle

• Angles: the rotational measure of a ray

• Degree measure: if the rotation from the first side to the last side is (1/360th) of rotation, the angle now measures 1 degree or 1°

• Radian measure: it is the measure of the angle subtended by an arc

• The relation between degree and radian:

• Notational conversions

• Degree measure = 180/pie ×radian measures

• Signs of Trigonometric functions: depending on the Trigonometric functions and their positions in the quadrant, they may be + or -

• Trigonometric identities: the result of the sum or difference of two numbers and their related expressions.

Trigonometric Equations

Variables of an equation with Trigonometric functions are called Trigonometric equations.

The solution of these equations is called principle solutions. If this equation involves integer n then its solution is called the general solution.

Z is used to denote a set of integers.

Points to remember before the exercise:

• sin x =0 gives x = nπ, where n ∈ Z

• cos x =0 gives x = (2n + 1) π/2, where n ∈ Z.

Theorems Discussed For Trigonometric Functions Exercise 3.4:

• Theorem 1: It states that for real numbers x and y sin x= sin y means x= nπ + (–1)n y, and n ∈ Z.

• Theorem 2: It states that for real numbers x and y, cos x = cos y, means x = 2nπ ± y, and n ∈ Z.

• Theorem 3: It states that if x and y are not odd multiples of π/2, then tan x = tan y means x = nπ + y, where n ∈ Z

Trigonometric Functions Exercise 3.4:

The solution of a total of 12 equations asked in exercise 3.4 chapter Trigonometric Functions is available to help the students with relevant and precise solutions asked in the textbook. This exercise is based on principle solutions and general solutions discussed above. Also, it involves the three Theorems listed.

Tips to Solve NCERT Solutions of Exercise 3.4 Class 11 Maths Chapter 3

Exercise 3.4 of NCERT  Class 11 Maths Chapter 3 contains 9 questions that deal with the concept of trigonometric equations, which is the most important part of the entire Chapter 3 Trigonometric Functions of Class 11 Maths. All the formulas and trigonometric identities you have learned in the previous exercises are to be applied here. Therefore, students need to develop the right skill set for solving trigonometric equations.

Here are some tips and points to remember if students want to navigate through Exercise 3.4 in Chapter 3 of Class 11 Maths smoothly.

• Go through Vedantu’s NCERT Solutions for Maths Class 11 Chapter 3 (Ex 3.4) thoroughly as these are well-formed with an easy-to-understand pattern to help understand the topic clearly and secure good marks in exams.

• Before starting to solve the questions of Chapter 3 of Class 11 Maths Exercise 3.4,  quickly go through the trigonometric terms, formulas, and identities, and remember them.

• Clear all the doubts as soon as possible.

• Always solve the trigonometric equations questions step-wise. You can first refer to the examples provided in NCERT Solutions for Class 11 Maths Chapter 3 by Vedantu.

NCERT Solution Class 11 Maths of Chapter 3 All Exercises

 Chapter 3 - Trigonometric Functions Exercises in PDF Format Exercise 3.1 7 Questions & Solutions Exercise 3.2 10 Questions & Solutions Exercise 3.3 25 Questions & Solutions Exercise 3.4 9 Questions & Solutions Miscellaneous Exercise 10 Questions & Solutions