Solve $1 + sin 2 x = (sin 3 x - cos 3 x)^{2}$ .

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Solution

$1 + sin 2 x = 1 - sin 6 x$
$∴ sin 6 x + sin 2 x = 0$
$∴ 2 sin 4 x cos 2 x = 0$
$∴ sin 4 x = 0$
$∴ x = n π / 4$
or $cos 2 x = 0$
$∴ 2 x = (n + \frac{1}{2}) π$
or $x = (n + \frac{1}{2}) \frac{π}{2}$ .

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