If sin 5x + sin 3 x + sin x = 0, then the value of x other than zero, lying between 0

Question

If sin 5x + sin 3 x + sin x = 0, then the value of x other than zero, lying between 0 <x<&pi;/2 isa)&pi;/6b)&pi;/12c)&pi;/3d)&pi;/4Correct answer is option 'C'. Can you explain this answer?

Answer

1. sin 5x + sin x + sin 3x = 0
2 sin 3x . cos2x + sin3x = 0
sin 3x (2 cos 2x + 1) = 0, 0 ≤ x ≤π/2

From both x = π / 3 (other than 0).

Answer

Sin5x+sin X +sin3x=0
2sin3x cos2x+sin3x=0. (sin c + sin d)
sin3x(2cos2x+1)=0
put sin 3x=0
X=nπ
(X=0,π,2π,....,nπ)
and also,
2cos2x+1=0
cos2x=-1/2
2x=2π/3
X=π/3
now ATQ,
X=π/3 answer

Answer

Sometimes its better to go by optional approach.
But if you still want to solve it by general method then use the formula
SinC+SinD= 2Sin{(C+D)/2} Cos{(C-D)/2}

Answer

We can rewrite the given equation as:

sin(5x) * sin(3x) * sin(x) = 0

Since the product of three factors is equal to zero, at least one of the factors must be equal to zero. Therefore, we have three possible cases:

Case 1: sin(5x) = 0

This occurs when 5x is an integer multiple of π. In other words:

5x = nπ

where n is an integer. Dividing both sides by 5 gives:

x = nπ/5

Case 2: sin(3x) = 0

This occurs when 3x is an integer multiple of π. In other words:

3x = mπ

where m is an integer. Dividing both sides by 3 gives:

x = mπ/3

Case 3: sin(x) = 0

This occurs when x is an integer multiple of π. In other words:

x = kπ

where k is an integer.

Therefore, the possible values of x, other than zero, lying between 0 and 2π are:

x = nπ/5, mπ/3, kπ

where n, m, and k are integers.

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