Ncert Solution 3.4 Class 11

Find the principal and general solutions of the following equations:

Question 1. tan x = √3

Solution:

Given: tan x = √3

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Here, x lies in first or third quadrant.


∴ tanx = tan60°∘ or tanx = tan(180° + 60°)

tanx = tan60°∘ or tanx = tan240°∘

Here, the principal solutions areπ/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 = cos π/3 or cosx = cos 5π/3

Here, the principal solutions are π/3, 5 π/3

∴ cosx = cosπ\frac{ π}{3}

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) n 7π/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 = 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 =  orx = 3​, n Z

Therefore, theprincipalsolutionsare, ​/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​ orcosx = 1/2 ​= cosπ/3​, n z

x = (2n + 1)π/4​ orx = 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​ orsinx = −1/2​ = −sinπ/6​, n z

x = (2n + 1)π/4​ orx = 2± 2π/3​, n z

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

x = (2n + 1)π/4​ orx = + (-1) n 7π/6​, n Z

Question 8. sec22x = 1 – tan 2x

Solution:

Given: sec22x = 1 – tan 2x

1 + tan 22x = 1 – tan2x

tan2x(tan2x + 1) = 0

tan2x = 0 ortan2x + 1 = 0

2x =  ortan2x – 1 = –tan π/4

x = nπ/2​ orx = ​/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 =  orcos2x = -1/2 ​= cos2π/3​, n z

x = nπ/3​ or 2x = nπ/2 ± 2π/3​, n z

x = ​/3 orx = ± π/3​, n z


Ncert Solution 3.4 Class 11

Source: https://www.geeksforgeeks.org/class-11-ncert-solutions-chapter-3-trigonometric-function-exercise-3-4/

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