# From the equation (sin2x)(sin6x)=(sinx)(sin7x), what is x?

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To solve this equation, we'll change the products into sums, using the next formula:

(sin a)*(sin b) = [cos(a-b)-cos(a+b)]/2

Let a = 2x and b = 6x:

(sin2x)*(sin6x) = [cos(2x-6x)-cos(2x+6x)]/2

(sin2x)*(sin6x) = [cos(-4x)-cos(8x)]/2

Let a = x and b = 7x:

(sinx)*(sin7x) = [cos(x-7x)-cos(x+7x)]/2

(sinx)*(sin7x) = [cos(-6x)-cos(8x)]/2

We'll re-write the equation:

[cos(-4x)-cos(8x)]/2 = [cos(-6x)-cos(8x)]/2

[cos(-4x)-cos(8x)] = [cos(-6x)-cos(8x)]

We'll eliminate cos(8x):

cos(-4x) = cos(-6x)

Since the cosine function is even, we'll get:

cos (4x) = cos (6x)

We'll subtract cos (6x) both sides:

cos (4x) - cos (6x) = 0

Since the trigonometric functions are matching, we'll transfrom the difference into a product, usinfg the formula:

cos a - cos b = -2sin [(a+b)/2]*sin[(a-b)/2]

cos 4x - cos 6x = -2sin [(4x+6x)/2]*sin[(4x-6x)/2]

cos 4x - cos 6x = 2(sin 5x)*(sin x)

We'll re-write the equation:

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

We'll cancel each factor:

sin 5x = 0

5x = (-1)^k*arcsin 0 + k*pi

5x = k*pi

x = k*pi/5

sin x = 0

x = k*pi

**The solutions of the equation are represented by the reunion of sets: {k*pi/5 / k is integer number}U{k*pi / k is integer number}.**