To solve this equation, we first notice that it is a quadratic equation in terms of sin 2x. Let's make a substitution to simplify the equation.

Let y = sin 2x.
Our equation becomes:
2y^2 - 2sin4x + 1 = 0.

Now, notice that sin 4x can be represented in terms of sin 2x using the double angle formula:
sin 4x = 2sin 2x cos 2x
Substitute this into the equation:
2y^2 - 2(2y)(√(1-y^2)) + 1 = 0
2y^2 - 4y√(1-y^2) + 1 = 0

Now, this is a quadratic equation in terms of y. Let's solve it by treating it as a quadratic equation.
Let u = √(1 - y^2):
2(y^2 + 1) - 4y*u + 1 = 0
2y^2 + 2 - 4y√(1-y^2) + 1 = 0
2y^2 - 4yu + 1 = 0

Now, this is a standard quadratic equation in terms of y. We can solve it using the quadratic formula:
y = (4u ± √(16u^2 - 8))/4

Substitute u back in terms of y:
u = √(1 - y^2)
y = (4√(1-y^2) ± √(16(1-y^2)^2 - 8))/4

Solve for y, then substitute back y to find x.