Sin3x-sin7x=(√3)*sin2x

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To solve this equation, we can start by using the sum-to-product trigonometric identity:

sin(a) - sin(b) = 2sin((a - b)/2)cos((a + b)/2)

Applying this identity to our equation, we have:

sin(3x) - sin(7x) = 2sin((3x - 7x)/2)cos((3x + 7x)/2)
sin(3x) - sin(7x) = 2sin(-2x/2)cos(5x/2)
sin(3x) - sin(7x) = 2sin(-x)cos(5x/2)
sin(3x) - sin(7x) = -2sin(x)cos(5x/2)

Now, our equation becomes:

-2sin(x)cos(5x/2) = √3*sin(2x)

Using the double angle identity sin(2x) = 2sin(x)cos(x), we have:

-2sin(x)cos(5x/2) = √3*2sin(x)cos(x)
-2sin(x)cos(5x/2) = 2√3sin(x)cos(x)

Dividing by 2sin(x) on both sides:

-cos(5x/2) = √3*cos(x)

Dividing by cos(x) on both sides:

-cot(5x/2) = √3

Taking the arccotangent of both sides:

5x/2 = arccot(-√3)

Finally, solving for x:

x = 2*(arccot(-√3))/5

Therefore, the solution to the equation sin(3x) - sin(7x) = √3sin(2x) is x = 2(arccot(-√3))/5.