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
Â
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 2Ï€/3
Here
