Class 11 Maths Ncert Solutions Trigonometry Video

Chapter 3 Trigonometric Functions of Class 11 Maths is categorized under the term – II CBSE Syllabus for 2021-22 and the solutions provided here are based on the latest guidelines prescribed by the CBSE. The NCERT Solutions for class 11 can help the students in preparing well for the Class 11 second term exams. These solutions will help students in understanding the problem-solving method and analyse the different types of questions that might arise in the term – II exams.

Here, Exercise 3.4 of NCERT Solutions for Class 11 Maths Chapter 3- Trigonometric Functions is based on trigonometric equations. Equations involving trigonometric functions of a variable are called trigonometric equations. Students will learn to solve the problems related to different trigonometric equations in this exercise.

Download PDF of NCERT Solutions for Class 11 Maths Chapter 3- Trigonometric Functions Exercise 3.4

Access other exercise solutions of Class 11 Maths Chapter 3- Trigonometric Functions

Exercise 3.1 Solutions 7 Questions

Exercise 3.2 Solutions 10 Questions

Exercise 3.3 Solutions 25 Questions

Miscellaneous Exercise On Chapter 3 Solutions 10 Questions

Access Solutions for Class 11 Maths Chapter 3.4 exercise

Find the principal and general solutions of the following equations:

1. tan x = √3

Solution:

2. sec x = 2

Solution:

3. cot x = – √3

Solution:

4. cosec x = – 2

Solution:

Find the general solution for each of the following equations:

5. cos 4x = cos 2x

Solution:

6. cos 3x + cos x – cos 2x = 0

Solution:

7. sin 2x + cos x = 0

Solution:

It is given that

sin 2x + cos x = 0

We can write it as

2 sin x cos x + cos x = 0

cos x (2 sin x + 1) = 0

cos x = 0 or 2 sin x + 1 = 0

Let cos x = 0

8. sec² 2x = 1 – tan 2x

Solution:

It is given that

sec² 2x = 1 – tan 2x

We can write it as

1 + tan² 2x = 1 – tan 2x

tan² 2x + tan 2x = 0

Taking common terms

tan 2x (tan 2x + 1) = 0

Here

tan 2x = 0 or tan 2x + 1 = 0

If tan 2x = 0

tan 2x = tan 0

We get

2x = nπ + 0, where n ∈ Z

x = nπ/2, where n ∈ Z

tan 2x + 1 = 0

We can write it as

tan 2x = – 1

So we get

Here

2x = nπ + 3π/4, where n ∈ Z

x = nπ/2 + 3π/8, where n ∈ Z

Hence, the general solution is nπ/2 or nπ/2 + 3π/8, n ∈ Z.

9. sin x + sin 3x + sin 5x = 0

Solution:

It is given that

sin x + sin 3x + sin 5x = 0

We can write it as

(sin x + sin 5x) + sin 3x = 0

Using the formula

By further calculation

2 sin 3x cos (-2x) + sin 3x = 0

It can be written as

2 sin 3x cos 2x + sin 3x = 0

By taking out the common terms

sin 3x (2 cos 2x + 1) = 0

Here

sin 3x = 0 or 2 cos 2x + 1 = 0

If sin 3x = 0

3x = nπ, where n ∈ Z

We get

x = nπ/3, where n ∈ Z

If 2 cos 2x + 1 = 0

cos 2x = – 1/2

By further simplification

= – cos π/3

= cos (π – π/3)

So we get

cos 2x = cos 2π/3

Here

Class 11 Maths Ncert Solutions Trigonometry Video

Source: https://byjus.com/ncert-solutions-class-11-maths-chapter-3-trigonometric-functions-ex-3-4/

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