Sin²(2x)=cos2x+4sin⁴x

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To solve this equation, we can use the trigonometric identity:

sin^2(2x) = 1 - cos^2(2x)

So our equation becomes:

1 - cos^2(2x) = cos(2x) + 4sin^4(x)

Now we can substitute cos(2x) = 1 - 2sin^2(x) and simplify:

1 - (1 - 2sin^2(x))^2 = (1 - 2sin^2(x)) + 4sin^4(x)

Expanding and simplifying:

1 - (1 - 4sin^2(x) + 4sin^4(x)) = 1 - 2sin^2(x) + 4sin^4(x)

1 - 1 + 4sin^2(x) - 4sin^4(x) = 1 - 2sin^2(x) + 4sin^4(x)

4sin^2(x) - 4sin^4(x) = 1 - 2sin^2(x) + 4sin^4(x)

Adding 2sin^2(x) and subtracting 4sin^2(x) from both sides:

6sin^2(x) = 1

Dividing by 6:

sin^2(x) = 1/6

Taking the square root of both sides:

sin(x) = ±√1/6

Therefore, the solutions for x are:

x = arcsin(√1/6) or x = π - arcsin(√1/6) (and any integer multiple of π)