# Class 11 NCERT Solutions- Chapter 3 Trigonometric Function – Exercise 3.4

### Question 1. tan x = âˆš3

Solution:

Given: tan x = âˆš3

Here, x lies in first or third quadrant.

âˆ´ tanx = tan60Â°âˆ˜ or tanx = tan(180Â° + 60Â°)

tanx = tan60Â°âˆ˜ or tanx = tan240Â°âˆ˜

Here, the principal solutions are Ï€/3â€‹, 4Ï€/3â€‹.

âˆ´ tanx = tanÏ€/3

â‡’ x = nÏ€ + Ï€â€‹/3 where n âˆˆ Zâ€‹

### 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.

âˆ´ cosx = cos60Â°âˆ˜ or cosx = cos(360Â° – 60Â°)

cosx = cos60Â°âˆ˜ or cosx = cos300Â°âˆ˜

cosx = cosÏ€/3 or cosx = cos 5Ï€/3

Here, the principal solutions are Ï€/3, 5 Ï€/3

âˆ´ cosx = cosÏ€

x = 2nÏ€ Â± Ï€/3â€‹ where, n âˆˆ Zâ€‹

### Question 3. cot x = âˆ’âˆš3

Solution:

Given: cot x = âˆ’âˆš3

It can be written as, tan x = -1/âˆš3

Here, x lies in second or fourth quadrant.

âˆ´ tanx = âˆ’tan30Â°âˆ˜= tan(180Â°– 30Â°) or tanx = tan(360Â°âˆ˜- 60Â°)

tanx = tan150Â°âˆ˜ or tanx = tan330Â°âˆ˜

tanx = tan5Ï€/6â€‹ or tanx = tan11Ï€/6â€‹â€‹

Here, the principal solutions are 5Ï€/6, 11 Ï€/6

âˆ´ tan x = tan 5Ï€/6

â‡’x = nÏ€ + 5Ï€/3â€‹ where, n âˆˆ Zâ€‹

### 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.

âˆ´ sinx = -sin30Â°âˆ˜= sin(180Â° + 30Â°) or sinx = sin(360Â°âˆ˜- 30Â°)

sinx = sin210Â°âˆ˜ or sinx = sin330Â°âˆ˜

sinx = sin7Ï€/6â€‹ or sinx = sin11Ï€/6â€‹â€‹

Here, the principal solutions are 7Ï€/6, 11 Ï€/6

âˆ´ sin x = -sin Ï€/6

â‡’ x = nÏ€ + (âˆ’1)n7Ï€/6â€‹ where, n âˆˆ Zâ€‹

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

Solution:

Given: cos 4x = cos 2x

4x = 2nÏ€ Â± 2x, n âˆˆ Z

4x – 2x = 2nÏ€ or 4x + 2x = 2nÏ€, n âˆˆ Z

2x = 2nÏ€ or 6x = 2nÏ€, n âˆˆ Z

2x = 2nÏ€ or 6x = 2nÏ€, n âˆˆ Z

x = nÏ€ or x = 3nÏ€â€‹, n âˆˆ Z

Therefore, the principal solutions are nÏ€, nÏ€â€‹/3

### Question 6. cos 3x + cos x â€“ cos 2x = 0

Solution:

Given: cos 3x + cos x â€“ cos 2x = 0

2cos2xcosx – cos2x = 0

cos2x(2cosx – 1) = 0

cos2x = 0 or 2cosx – 1 = 0

2x = (2n + 1)Ï€/2â€‹ or cosx = 1/2 â€‹= cosÏ€/3â€‹, n âˆˆ z

x = (2n + 1)Ï€/4â€‹ or x = 2nÏ€ Â± 2Ï€â€‹/3, n âˆˆ 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)Ï€/2â€‹ or sinx = âˆ’1/2â€‹ = âˆ’sinÏ€/6â€‹, n âˆˆ z

x = (2n + 1)Ï€/4â€‹ or x = 2nÏ€ Â± 2Ï€/3â€‹, n âˆˆ z

x = (2n + 1)Ï€/4â€‹ or x = nÏ€ + (-1)n-Ï€/6â€‹, n âˆˆ Z

x = (2n + 1)Ï€/4â€‹ or x = nÏ€ + (-1)n7Ï€/6â€‹, n âˆˆ Z

### Question 8. sec22x = 1 â€“ tan 2x

Solution:

Given: sec22x = 1 – tan 2x

1 + tan22x = 1 – tan2x

tan2x(tan2x + 1) = 0

tan2x = 0 or tan2x + 1 = 0

2x = nÏ€ or tan2x – 1 = –tanÏ€/4

x = nÏ€/2â€‹ or x = nÏ€â€‹/2 + 3Ï€/8â€‹, n = Zâ€‹

### 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 = nÏ€ or cos2x = -1/2 â€‹= cos2Ï€/3â€‹, n âˆˆ z

x = nÏ€/3â€‹ or 2x = nÏ€/2 Â± 2Ï€/3â€‹, n âˆˆ z

x = nÏ€â€‹/3 or x = nÏ€ Â± Ï€/3â€‹, n âˆˆ z

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