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Class 11 NCERT Solutions- Chapter 3 Trigonometric Function – Exercise 3.4
• Last Updated : 02 Feb, 2021

### 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

x = 2± 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, the principal solutions are 5π/6, 11 π/6

∴ tan x = tan 5π/6

x = + 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 = + (−1)n7π/6​ where, n Z

### Question 5. cos 4x = cos 2x

Solution:

Given: cos 4x = cos 2x

4x = 2± 2x, n Z

4x – 2x = 2 or 4x + 2x = 2, n Z

2x = 2 or 6x = 2, n Z

2x = 2 or 6x = 2, n Z

x =  or x = 3​, n Z

Therefore, the principal solutions are , ​/3

### Question 6. cos 3x + cos x – cos 2x = 0

Solution:

Given: cos 3x + cos x – cos 2x = 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 = 2± 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 = 2± 2π/3​, n z

x = (2n + 1)π/4​ or x = + (-1)n-π/6​, n Z

x = (2n + 1)π/4​ or x = + (-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 =  or tan2x – 1 = –tanπ/4

x = nπ/2​ or x = ​/2 + 3π/8​, n = Z

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

Solution:

Given: sin x + sin 3x + sin 5x = 0

2sin3xcos2x + sin3x = 0

sin3x(2cos2x + 1) = 0

sin3x = 0 or 2cos2x + 1 = 1

3x =  or cos2x = -1/2 ​= cos2π/3​, n z

x = nπ/3​ or 2x = nπ/2 ± 2π/3​, n z

x = ​/3 or x = ± π/3​, n z

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