How do you solve the following equation #[sin (2x) + cos (2x)] ^2 = 1# in the interval [0, 2pi]?
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To solve the equation ([sin(2x) + cos(2x)]^2 = 1) in the interval ([0, 2\pi]), follow these steps:

Expand the square: ([sin(2x) + cos(2x)]^2 = sin^2(2x) + 2sin(2x)cos(2x) + cos^2(2x))

Use the trigonometric identity (sin^2(\theta) + cos^2(\theta) = 1) to simplify: (sin^2(2x) + 2sin(2x)cos(2x) + cos^2(2x) = 1 + 2sin(2x)cos(2x))

Replace (sin(2x)) and (cos(2x)) with their respective identities: (1 + 2sin(2x)cos(2x) = 1 + 2[2sin(x)cos(x)][cos^2(x)  sin^2(x)])

Simplify further: (1 + 2[2sin(x)cos(x)][cos^2(x)  sin^2(x)] = 1 + 2[2sin(x)cos(x)][cos^2(x)  (1  cos^2(x))])

Expand and simplify the expression: (1 + 2[2sin(x)cos(x)][cos^2(x)  (1  cos^2(x))] = 1 + 2[2sin(x)cos(x)][2cos^2(x)  1])

Expand and simplify again: (1 + 2[2sin(x)cos(x)][2cos^2(x)  1] = 1 + 8sin(x)cos(x)cos^2(x)  4sin(x)cos(x))

Use the identity (sin(2\theta) = 2sin(\theta)cos(\theta)): (1 + 8sin(x)cos(x)cos^2(x)  4sin(x)cos(x) = 1 + 4sin(2x)cos^2(x)  2sin(2x))

Replace (sin(2x)) with (2sin(x)cos(x)): (1 + 4sin(2x)cos^2(x)  2sin(2x) = 1 + 8sin(x)cos(x)cos^2(x)  2[2sin(x)cos(x)])

Simplify: (1 + 8sin(x)cos(x)cos^2(x)  2[2sin(x)cos(x)] = 1 + 8sin(x)cos^3(x)  4sin(x)cos(x))

Rearrange the terms: (1 + 8sin(x)cos^3(x)  4sin(x)cos(x)  1 = 0)

Combine like terms: (8sin(x)cos^3(x)  4sin(x)cos(x) = 0)

Factor out (4sin(x)cos(x)): (4sin(x)cos(x)(2cos^2(x)  1) = 0)

Solve each factor separately: (4sin(x)cos(x) = 0) or (2cos^2(x)  1 = 0)

Solve (4sin(x)cos(x) = 0) for (x): (sin(x) = 0) or (cos(x) = 0)

For (sin(x) = 0), (x = 0, \pi). For (cos(x) = 0), (x = \frac{\pi}{2}, \frac{3\pi}{2}).

Solve (2cos^2(x)  1 = 0) for (x): (cos^2(x) = \frac{1}{2})

(cos(x) = \pm \frac{\sqrt{2}}{2})

For (cos(x) = \frac{\sqrt{2}}{2}), (x = \frac{\pi}{4}, \frac{7\pi}{4}). For (cos(x) = \frac{\sqrt{2}}{2}), (x = \frac{3\pi}{4}, \frac{5\pi}{4}).
Thus, the solutions in the interval ([0, 2\pi]) are (x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}).
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