$sinx + cosx.sin2x + \sqrt3 cos3x = 2(cos4x + sin^3x)$
$\Leftrightarrow sinx – 2sin^3x + cosx.sin2x + \sqrt3 cos3x = 2cos4x $
$\Leftrightarrow sinx.cos2x +cosx.sin2x + \sqrt3 cos3x = 2cos4x $
$\Leftrightarrow sin3x + \sqrt3 cos3x = 2cos4x $
$\Leftrightarrow \frac{1}2 sin3x + \frac{\sqrt3}2 cos3x = cos4x$
$\Leftrightarrow cos(3x –\frac{\pi}6) = cos4x$
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