Sin4x = 2cos^2(x + pi/3) - 1

To solve the equation sin(4x) = 2cos^2(x + π/3) - 1, we can use the trigonometric identity sin(2θ) = 2sin(θ)cos(θ).

First, let's simplify the right side of the equation using the given identity:
2cos^2(x + π/3) - 1 = 2(cos(x + π/3))^2 - 1
= 2(cos^2(x)cos^2(π/3) - sin^2(x)sin^2(π/3)) - 1
= 2(cos^2(x)(1/4) - sin^2(x)(3/4)) - 1
= (1/2)cos^2(x) - (3/2)sin^2(x) - 1

Now, substitute this expression back into the original equation:
sin(4x) = (1/2)cos^2(x) - (3/2)sin^2(x) - 1

Using the double angle identity sin(2θ) = 2sin(θ)cos(θ), we can rewrite sin(4x) as:
sin(4x) = 2sin(2x)cos(2x)

Now, substitute this back into the equation:
2sin(2x)cos(2x) = (1/2)cos^2(x) - (3/2)sin^2(x) - 1

Next, let's use the double angle identity for cosine: cos(2θ) = cos^2(θ) - sin^2(θ)
2sin(2x)cos(2x) = (1/2)(cos(2x) + 1) - (3/2)sin^2(x) - 1

Distribute the (1/2) to the terms inside the parentheses:
2sin(2x)cos(2x) = (1/2)cos(2x) + 1/2 - (3/2)sin^2(x) - 1

Combine like terms:
2sin(2x)cos(2x) = (1/2)cos(2x) - (3/2)sin^2(x) - 1/2

Multiply both sides of the equation by 2 to eliminate the fractions:
4sin(2x)cos(2x) = cos(2x) - 3sin^2(x) - 1

Now, let's simplify the equation further:
4sin(2x)cos(2x) - cos(2x) + 3sin^2(x) + 1 = 0

Factor out a common factor of cos(2x):
cos(2x)(4sin(2x) - 1) + 3sin^2(x) + 1 = 0

Now, let's focus on the first term:
cos(2x)(4sin(2x) - 1)

To find the values of x that satisfy this term, we have two cases to consider:

Case 1: cos(2x) = 0
This occurs when 2x = π/2 + nπ or 2x = 3π/2 + nπ, where n is an integer.
Solving for x, we have x = π/4 + (nπ)/2 or x = (3π/4 + nπ)/2, where n is an integer.

Case 2: 4sin(2x) - 1 = 0
Solving for sin(2x), we have sin(2x) = 1/4
This occurs when 2x = π/6 + 2nπ or 2x = 5π/6 + 2nπ, where n is an integer.
Solving for x, we have x = π/12 + (nπ)/2 or x = (5π/12 + nπ)/2, where n is an integer.

Therefore, the solutions to the equation sin(4x) = 2cos^2(x + π/3) - 1 are:
x = π/4 + (nπ)/2, (3π/4 + nπ)/2, π/12 + (nπ)/2, (5π/12 + nπ)/2, where n is an integer.