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

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Find the principal and general solutions of the 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 Ï€/3​, 4Ï€/3​.

∴ tanx = tanÏ€/3

⇒ x = nÏ€ + 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 = cosÏ€\frac{ π}{3}

x = 2nÏ€ ± Ï€/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 = tan5Ï€/6​ or tanx = tan11Ï€/6​​

Here, the principal solutions are 5π/6, 11 π/6

∴ tan x = tan 5π/6

⇒x = nÏ€ + 5Ï€/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 = sin7Ï€/6​ or sinx = sin11Ï€/6​​

Here, the principal solutions are 7π/6, 11 π/6

∴ sin x = -sin π/6

⇒ x = nÏ€ + (−1)n7Ï€/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 = 2nÏ€ or 4x + 2x = 2nÏ€, n ∈ Z

2x = 2nÏ€ or 6x = 2nÏ€, n ∈ Z

2x = 2nÏ€ or 6x = 2nÏ€, n ∈ Z

x = nÏ€ or x = 3nπ​, n ∈ Z

Therefore, the principal solutions are nÏ€, 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)Ï€/2​ or cosx = 1/2 ​= cosÏ€/3​, n ∈ z

x = (2n + 1)Ï€/4​ or x = 2nÏ€ ± 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)Ï€/2​ or sinx = −1/2​ = −sinÏ€/6​, n ∈ z

x = (2n + 1)Ï€/4​ or x = 2nÏ€ ± 2Ï€/3​, n ∈ z

x = (2n + 1)Ï€/4​ or x = nÏ€ + (-1)n-Ï€/6​, n ∈ Z

x = (2n + 1)Ï€/4​ or x = nÏ€ + (-1)n7Ï€/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 = nÏ€ or tan2x – 1 = –tanÏ€/4

x = nÏ€/2​ or x = nπ​/2 + 3Ï€/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 = nÏ€ or cos2x = -1/2 ​= cos2Ï€/3​, n ∈ z

x = nÏ€/3​ or 2x = nÏ€/2 ± 2Ï€/3​, n ∈ z

x = nπ​/3 or x = nÏ€ ± Ï€/3​, n ∈ z


Last Updated : 02 Feb, 2021
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