Answer
$\cos x(1-4\sin^{2}x)$
(equivalent answers may differ if we choose other double angle identities for cosine)
Work Step by Step
$\cos 3x=\cos(x+2x)=$
... Sum identity:
$... \cos(A+B)=\cos A\cos B-\sin A\sin B$
$= \cos x\cos 2x-\sin x\sin 2x$
$...$Double angle identities:
$...\sin 2A=2\sin A\cos A$
$...\cos 2A=1-2\sin^{2}A$
$=\cos x(1-2\sin^{2}x)-\sin x(2\sin x\cos x)$
$=\cos x-2\sin^{2}x\cos x-2\sin^{2}x\cos x$
$=\cos x-4\sin^{2}x\cos x$
$=\cos x(1-4\sin^{2}x)$
(answers may differ if we choose other double angle identities for cosine,
$\cos 2A=\cos^{2}A-\sin^{2}A$,
$\cos 2A=2\cos^{2}A- \mathrm{l}$)
