$\Leftrightarrow(\frac{1-\cos 2x}{2})^2+(\frac{1+\sin 2x}{2})^2+(\frac{1-\sin 2x}{2})^2=\frac{9}{8}$
$\Leftrightarrow\frac{4+\sin^2 2x-2\cos 2x}{4}=\frac{9}{8}$
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