$\left \{ {{x = (2k + 1)π/2} \atop {x = - π/8 + kπ}} \atop {x = 5π/16 + kπ/2}\right.$
Giải thích các bước giải:
$Sin2x - cos2x = 1 + √2(sin2x + sin4x)$
$⇔ sin2x - (1 + cos2x) - √2(sin2x + sin4x) = 0$
$⇔ 2sinxcosx - 2cos²x - 2√2sin3xcosx = 0$
$⇔ 2cosx(sinx - cosx - √2sin3x) = 0$
$⇔ 2cosx.[√2sin(x - π/4) - √2sin3x] = 0$
$⇔ 2√2cosx[sin(x - π/4) - sin3x] = 0$
$⇔ 4√2cosx.cos(2x - π/8)sin = 0$
$⇔ sin3x = sin(x - π/4)$
$⇔\left \{ {{3x = x - π/4 + 2kπ} \atop {3x = π - (x - π/4) + 2kπ}} \right.$
$⇔\left \{ {{x = - π/8 + kπ} \atop {x = 5π/16 + kπ/2}} \right.$
