NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions Exercise 3.4" provides comprehensive answers and explanations to the questions and problems presented in Exercise 3.4 of the textbook.
This exercise focuses on exploring the properties and applications of trigonometric functions, including sine, cosine, and tangent, as well as their inverses. By following these solutions, students can deepen their understanding of trigonometry and enhance their problem-solving skills.
Solve Following Questions NCERT Solutions for Class 12 Math's Chapter 1 (Trigonometric Function) - Exercise3.4
Find the Principal and General Solutions of Following Equations:
Question 1. tan x = √3
Solution:
Given: tan x = √3
Here, x lies in first or third quadrant.
∴ tanx = tan60°∘ or tanx = tan(180° + 60°)
tanx = tan60°∘ or tanx = tan240°∘
Here, the principal solutions are π/3, 4π/3.
∴ tanx = tanπ/3
⇒ x = nπ + π/3 where n ∈ Z
Question 2. sec x = 2
Solution:
Given: sec x = 2
It can be written as, cos x = 1/2
Here, x lies in first or fourth quadrant.
∴ cosx = cos60°∘ or cosx = cos(360° - 60°)
cosx = cos60°∘ or cosx = cos300°∘
cosx = cosπ/3 or cosx = cos 5π/3
Here, the principal solutions are π/3, 5 π/3
∴ cosx = cosπ
\frac{ π}{3} x = 2nπ ± π/3 where, n ∈ Z
Question 3. cot x = −√3
Solution:
Given: cot x = −√3
It can be written as, tan x = -1/√3
Here, x lies in second or fourth quadrant.
∴ tanx = −tan30°∘= tan(180°- 30°) or tanx = tan(360°∘- 60°)
tanx = tan150°∘ or tanx = tan330°∘
tanx = tan5π/6 or tanx = tan11π/6
Here, principal solutions are 5π/6, 11 π/6
∴ tan x = tan 5π/6
⇒x = nπ + 5π/3 where, n ∈ Z
Question 4. cosec x = - 2
Solution:
Given: cosec x = - 2
It can be written as, sin x = -1/2
Here x lies in third or fourth quadrant.
∴ sinx = -sin30°∘= sin(180° + 30°) or sinx = sin(360°∘- 30°)
sinx = sin210°∘ or sinx = sin330°∘
sinx = sin7π/6 or sinx = sin11π/6
Here, the principal solutions are 7π/6, 11 π/6
∴ sin x = -sin π/6
⇒ x = nπ + (−1)n7π/6 where, n ∈ Z
Find the General Solution for Each of the Following Equations:
Question 5. cos 4x = cos 2x
Solution:
Given: cos 4x = cos 2x
4x = 2nπ ± 2x, n ∈ Z
4x - 2x = 2nπ or 4x + 2x = 2nπ, n ∈ Z
2x = 2nπ or 6x = 2nπ, n ∈ Z
2x = 2nπ or 6x = 2nπ, n ∈ Z
x = nπ or x = 3nπ, n ∈ Z
Therefore, the principal solutions are nπ, nπ/3
Question 6. cos 3x + cos x – cos 2x = 0
Solution:
Given: cos 3x + cos x – cos 2x = 0
2cos\Big(\frac{3x+x}{2} \Big)cos\Big(\frac{3x-x}{2} \Big)-cos2x=0 2cos2xcosx - cos2x = 0
cos2x(2cosx - 1) = 0
cos2x = 0 or 2cosx - 1 = 0
2x = (2n + 1)π/2 or cosx = 1/2 = cosπ/3, n ∈ z
x = (2n + 1)π/4 or x = 2nπ ± 2π/3, n ∈ z
Question 7. sin 2x + cos x = 0
Solution:
Given: sin 2x + cos x = 0
2sinxcosx + cosx = 0
cosx(2sinx + 1) = 0
cosx = 0 or 2sinx + 1 = 0
x = (2n + 1)π/2 or sinx = −1/2 = −sinπ/6, n ∈ z
x = (2n + 1)π/4 or x = 2nπ ± 2π/3, n ∈ z
x = (2n + 1)π/4 or x = nπ + (-1)n-π/6, n ∈ Z
x = (2n + 1)π/4 or x = nπ + (-1)n7π/6, n ∈ Z
Question 8. sec22x = 1 – tan 2x
Solution:
Given: sec22x = 1 - tan 2x
1 + tan22x = 1 - tan2x
tan2x(tan2x + 1) = 0
tan2x = 0 or tan2x + 1 = 0
2x = nπ or tan2x - 1 = -tanπ/4
x = nπ/2 or x = nπ/2 + 3π/8, n = Z
Question 9. sin x + sin 3x + sin 5x = 0
Solution:
Given: sin x + sin 3x + sin 5x = 0
2sin\Big(\frac{5x+x}{2} \Big)cos\Big(\frac{5x-x}{2} \Big)+sin3x=0 2sin3xcos2x + sin3x = 0
sin3x(2cos2x + 1) = 0
sin3x = 0 or 2cos2x + 1 = 1
3x = nπ or cos2x = -1/2 = cos2π/3, n ∈ z
x = nπ/3 or 2x = nπ/2 ± 2π/3, n ∈ z
x = nπ/3 or x = nπ ± π/3, n ∈ z
Trigonometric Functions Exercises:
Class 11 NCERT Solutions – Chapter 3 Trigonometric Function – Exercise 3.2 |
|---|
Class 11 NCERT Solutions- Chapter 3 Trigonometric Function – Exercise 3.3 | Set 2 |