Math, asked by Anonymous, 1 year ago

If:
x^5 = 1
find the values of x.
There will be exactly 5 values of x.
Find it with full explanation for a 10th standard student only.
I will give 20 points for it.
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Answers

Answered by siddhartharao77
1

Step-by-step explanation:

Given : x⁵ = 1.

⇒ x = (1)^1/5.

Now, 1 can also be written as,

⇒ 1 = cos 2π + sin 2π.

It can also be written as,

⇒ 1 = cos 2nπ + sin 2nπ

using De Moivre's theorem, we get

⇒ (1)^1/5 = cos(2nπ/5) + i sin(2nπ/5).

The solutions of x⁵ - 1 = 0 can be obtained by putting n = 0,1,2,3,4.

(i) cos(0) + i sin(0) = 1.

(ii) cos(2π/5) + i sin(2π/5)  

(iii) cos(4π/5) + i sin(4π/5)

(iv) cos(6π/5) + i sin(6π/5)

(v) cos(8π/5) + i sin(8π/5)


Hope it helps!


Anonymous: what was stated in de moivre's theorem?
siddhartharao77: (cosθ + i sinθ)^n = cos nθ + i sin nθ.
Anonymous: oh thanks
siddhartharao77: wlcm
Answered by Ritiksuglan
0

Answer:

Step-by-step explanation:

Given : x⁵ = 1.

⇒ x = (1)^1/5.

Now, 1 can also be written as,

⇒ 1 = cos 2π + sin 2π.

It can also be written as,

⇒ 1 = cos 2nπ + sin 2nπ

using De Moivre's theorem, we get

⇒ (1)^1/5 = cos(2nπ/5) + i sin(2nπ/5).

The solutions of x⁵ - 1 = 0 can be obtained by putting n = 0,1,2,3,4.

(i) cos(0) + i sin(0) = 1.

(ii) cos(2π/5) + i sin(2π/5)  

(iii) cos(4π/5) + i sin(4π/5)

(iv) cos(6π/5) + i sin(6π/5)

(v) cos(8π/5) + i sin(8π/5)

Hope it helps!

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