Đáp án:
$x=\dfrac{k\pi}{2}$
Giải thích các bước giải:
$cos^3(3x)-cos^3x=4sin^2xcosx$
$\rightarrow cos^33x=cos^3x+4(1-cos^2x)cosx$
$\rightarrow cos^33x=cos^3x+4cosx-4cos^3x$
$\rightarrow cos^33x=cos^3x+cosx-(4cos^3x-3cosx)$
$\rightarrow cos^33x=cos^3x+cosx-cos3x$
$\rightarrow (cos^33x-cos^3x)+(cos3x-cosx)=0$
$\rightarrow (cos3x-cosx)(cos^23x+cos3x.cosx+cos^2x+1)=0$
$\rightarrow cos3x-cosx=0$
$\rightarrow cos3x=cosx$
$\rightarrow 3x=x+k2\pi\rightarrow x=k\pi$
Or $3x=-x+k2\pi\rightarrow x=\dfrac{k\pi}{2}$
$\Rightarrow x=\dfrac{k\pi}{2}$
