Đáp án: $x=k\pi$
Giải thích các bước giải:
Ta có:
$sin5x=5\sin x$
$\to \sin5x-\sin x=4\sin x$
$\to 2\cos(\dfrac{5x+x}{2})\cdot \sin(\dfrac{5x-x}{2})=4\sin x$
$\to 2\cos3x\sin2x=4\sin x$
$\to \cos3x\sin2x=2\sin x$
$\to \cos3x\cdot 2\sin x\cos x=2\sin x$
$\to \cos3x\cdot \sin x\cos x=\sin x$
$\to \cos3x\cdot \sin x\cos x-\sin x=0$
$\to \sin x(\cos3x\cdot \cos x-1)=0$
$\to \sin x=0\to x=k\pi$
Hoặc $\cos3x\cos x-1=0$
$\to \cos3x\cdot\cos x=1$
$\to \dfrac12(\cos(3x+x)+\cos(3x-x))=1$
$\to \cos4x+\cos2x=2$
Mà $\cos4x,\cos2x\le 1$
$\to \cos4x+\cos2x\le 2$
Dấu = xảy ra khi
$\cos4x=\cos2x=1$
$\to x=k\pi$
