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

NCERT Solutions of Chapter 3 Trigonometric Functions Exercise 3.4 will help you a lot if you want to excel in the examinations. The NCERT Solutions for Class 11 Maths prepared by Studyrankers experts are detailed and accurate so you can clear your doubts easily. You must try to understand the basic facts and important points before trying for advance problems and these NCERT Solutions will help you in developing these skillset.

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

1. (For Qs. 1-4): Find the principal and general solutions of the following equations:
tanx = √3

Answer

tan x = √3 = tan 60 °
∴ Principal value of x = 60° = π/3 radians and
= 4π/3
If tanθ = tan α where ‘α’ is the principal value of θ.
Then θ = nπ+ α
∴ General value of x = nπ + π/3, n  ∈ z.

2. sec x = 2

Answer

sec x = 2 = sec 60 or cos x = 1/2= cos 60
∴ Principal value = 60 = π/3 radians.
For cosθ = cosα , θ = 2nπ ± α
General value of x = 2nπ ± π/3

3. cot x = -√3

Answer

cotx = -√3, tan x = - 1/√3
= tan(180 - 30) = tan(π - π/6) = tan 5π/6
=> x = 5π/6
Principal solution = 5π/6, 11π/6
Genral solution = nπ ± 5π/6, n  ∈ z.

4. cosec x = -2

Answer

cosec θ = -2 or sin θ = -1/2
sin30 = 1/2 or sin(-30) = -sin30 = -1/2
Principal value of x = -30 = - π/6
general value of x = nπ +(-1)nα
= nπ+(-1)n(- π/6) = nπ -(-1)2(π/6)

5. Find the general solution of the following equations:
cos 4x = cos 2x

Answer

cos 4x = cos 2x
4x = 2nπ ± 2x
Taking + ve sign, we get
4x = 2nπ + 2x
=> 4x + 2x = 2nπ
=> 6x = 2nπ
x = nπ/3
=> x = nπ, n ∈ z

Taking -ve sign
4x = 2nπ - 2x
=> 3x + 2x = 2nπ
=> 6x = 2nπ
=> x = nπ/3, n  ∈ z
General solution is x = nπ/3 or x = nπ, n ∈ z

6. Find the general solution of the following equations :
Cos3x + cosx – cos2x = 0

Answer

Cos3x + cosx – cos2x = 0
Or 2cos (3x + x)/2 cos(3x – x)/2 – cos2x = 0
Or 2cos2xcosx – cos2x = 0
Or cos2x(2cosx – 1)= 0
If cos 2x = 0,
2x = (2n + 1) π/2 => x = (2n + 1) π/4
If 2cos x – 1 = 0, cos x = 1/2 => cosx = cos π/3
=> x = 2nπ ± π/3

7. Find the general solution of the following equations:
sin2x + cosx = 0

Answer

sin 2x + cos x = 0
=> 2cos x cos x cos x = 0
 => cos x (2 sin x + 1) = 0
=> cos x = 0
Or 2 sin x + 1 = 0
=> cos x = 1
Or sin x = -1/2 => cos x = 0
or sin x = sin( π + π/6) => cos x = 0
Or sin x = sin 7π/6 => x = (2n + 1)π/2
Or x = nπ +(-1)n 7π/6, n ∈ Z
Hence, general solution is
X = (2n + 1) π/2 or x = nπ +(-1)n 7π/6
where n ∈ Z

8. Find the general solution of
Sec22x = 1 – tan 2x

Answer

Sec22x = 1 – tan 2x
=> 1 + tan2 2x = 1 – tan 2x = 0
=> tan22x + tan2x = 0
=> tan 2x(tan 2x + 1) = 0
If tan 2x = 0, 2x = nπ or x = nπ/2
If tan 2x + 1 = 0, tan 2x = -1
= tan(π - π/4) = tan 3π/4
=> 2x = nπ + 3π/4 or x = nπ/2 + 3π/8

9. Find the general solution of 
sin x + sin3x + sin 5x = 0

Answer

We have, (sin 5x + sin x) + sin3x = 0
=> 2sin ((5x + x)/2) cos((5x - x)/2) + sin3x = 0
or 2sin3xcos2x + sin3x = 0
or sin 3x(2cos 2x + 1) = 0
if sin 3x = 0 => 3x = nπ or x = nπ/3
if 2cos2x + 1 = 0,
cos2x = -1/2 = cos(π - π/3) = cos 2π/3
∴ 2x = 2nπ ± 2π/3 or x = nπ ± π/3
Previous Post Next Post