4sin[tex]^{3}[/tex]x + 3cos[tex]^{3}[/tex]x +3= 6cos[tex]^{2}[/tex]( [tex]\frac{pi}{4}[/tex] - [tex]\frac{x}{2}[/tex]) + sin[tex]^{2}[/tex]x.cosx
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$4\sin^3 x+ 3\cos^3 x +3=3.[\sqrt{2} \cos (\dfrac{\pi}{4}-\dfrac{x}{2})]^2+\sin^2 x \cos x$
$\iff 4\sin^3 x+ 3\cos^3 x +3=3(\sin \dfrac{x}{2}+\cos \dfrac{x}{2} )^2+\sin^2 x. \cos x$
$\iff 4\sin^3 x+ 3\cos^3 x +3=3(1+\sin x)+\sin^2 x.\cos x$
$\iff 4\sin^3 x+ 3\cos^3 x +3=3+3\sin x+\sin^2 x.\cos x$
$\iff 4\sin^3 x+ 3\cos^3 x -3\sin x(\sin^2 x+\cos^2 x)-\sin^2 x.\cos x=0$
$\iff \sin^3 x-\sin^2 x.\cos x-3\sin x.\cos^2 x+3\cos^3 x=0$
$\iff (\sin x-\cos x)(\sin x -\sqrt{3}\cos x)(\sin x +\sqrt{3}\cos x)=0$
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