Đáp án: $x\in\{\dfrac14k\pi,\dfrac14(\pm\arccos(-\dfrac18)+k2\pi)\}$$
Giải thích các bước giải:
Ta có:
$\sin^3x\cos x-\cos^3x\sin x=\sin8x$
$\to \sin x\cos x(\sin^2x-\cos^2x)=\sin8x$
$\to \sin x\cos x\cdot (-\cos 2x)=\sin8x$
$\to -\sin x\cos x\cdot \cos 2x=\sin8x$
$\to -\dfrac12\cdot 2\sin x\cos x\cdot \cos 2x=\sin8x$
$\to -\dfrac12\cdot \sin2x\cdot \cos 2x=\sin8x$
$\to -\dfrac14\cdot 2\cdot \sin2x\cdot \cos 2x=\sin8x$
$\to -\dfrac14\cdot \sin4x=2\sin4x\cos4x$
$\to 2\sin4x\cos4x+\dfrac14\cdot \sin4x=0$
$\to 2\sin4x(\cos4x+\dfrac18)=0$
$\to\sin4x=0\to 4x=k\pi\to x=\dfrac14k\pi$
Hoặc $\cos4x+\dfrac18=0$
$\to \cos4x=-\dfrac18$
$\to 4x=\pm\arccos(-\dfrac18)+k2\pi$
$\to x=\dfrac14(\pm\arccos(-\dfrac18)+k2\pi)$
