2 sin^2 2x - 2 sin 4x + 1= 0

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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 2
Substitute this into the equation
2y^2 - 2(2y)(√(1-y^2)) + 1 =
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 =
2y^2 + 2 - 4y√(1-y^2) + 1 =
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.