2 cos4x + 2 cos2x = 4 sin^2 x -1

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To simplify this trigonometric equation, we can use the double angle formula for cosine, which states that cos(2θ) = 2cos^2(θ) - 1.

Therefore, we can rewrite the equation as follows:

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

Expanding and simplifying:

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

Now, we can use the double angle formula for sine, which states that sin(2θ) = 2sin(θ)cos(θ).

We will replace sin^2(x) with (1 - cos^2(x)):

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

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

4cos^2(2x) - 2 + 2cos(2x) = 3 - 4cos^2(x)

This simplifies to:

4cos^2(2x) + 2cos(2x) + 4cos^2(x) - 4 = 0

Factor the left side of the equation:
(2cos(2x) + 1)(2cos(2x) - 4cos(x) + 4) = 0

Now we have two possibilities:
1) 2cos(2x) + 1 = 0
2) 2cos(2x) - 4cos(x) + 4 = 0

Solving the first equation:
2cos(2x) + 1 = 0
2cos(2x) = -1
cos(2x) = -1/2

This means that 2x = 2π/3 or 2x = 4π/3

Therefore, x = π/3 or x = 2π/3

Solving the second equation:
2cos(2x) - 4cos(x) + 4 = 0
2(2cos^2(x) - 1) - 4cos(x) + 4 = 0
4cos^2(x) - 2 - 4cos(x) + 4 = 0
4cos^2(x) - 4cos(x) + 2 = 0

This is a quadratic equation in cos(x). By solving it, we get the values of x.

Therefore, the solutions to the equation 2cos4x + 2cos2x = 4sin^2 x - 1 are x = π/3, x = 2π/3, and the roots of the quadratic equation above for cos(x).