First, let's simplify the given equation:

sin6x + sin2x + 2sin"x = 1

Since we know that sin(2x) = 2sin(x)cos(x), we can substitute sin(2x) with 2sin(x)cos(x) in the equation:

sin6x + 2sin(x)cos(x) + 2sin(x) = 1

Now, let's convert all the terms to sin(x) terms using trigonometric identities:

sin6x = 2sin3xcos3x
2sin(x)cos(x) = sin2x
2sin(x) = 2sin(x)

Substituting these back into the equation, we get:

2sin3xcos3x + sin2x + 2sin(x) = 1

Now, let's square both sides of the equation to get rid of the square root:

(2sin3xcos3x + sin2x + 2sin(x))^2 = 1^2
(2sin3xcos3x + sin2x + 2sin(x))(2sin3xcos3x + sin2x + 2sin(x)) = 1
4sin^2(3x)cos^2(3x) + sin^2(2x) + 4sin^2(x) + 4sin(3x)cos(3x)sin(2x) + 2sin(3x)cos(3x)sin(2x) + 4sin(3x)cos(3x)sin(x) + sin(2x)2sin(x) + 2sin(x)sin(2x) + 4sin(x)sin(3x)cos(3x) = 1

This is the squared equation of the given equation: sin6x + sin2x + 2sin"x = 1".