Đáp án: $x\in\{\dfrac14\pi+k\pi,\dfrac34\pi+k\pi\}$
Giải thích các bước giải:
Ta có:
$\sin^6x+\cos^6x=\dfrac14\sin^22x$
$\to (\sin^2x)^3+(\cos^2x)^3=\dfrac14\sin^22x$
$\to (\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)=\dfrac14\sin^22x$
$\to 1-3\sin^2x\cos^2x\cdot 1=\dfrac14\sin^22x$
$\to 1-3\sin^2x\cos^2x=\dfrac14\sin^22x$
$\to 4-4\cdot 3\sin^2x\cos^2x=\sin^22x$
$\to 4- 3(2\sin x\cos x)^2=\sin^22x$
$\to 4- 3\sin^22x=\sin^22x$
$\to 4\sin^22x=4$
$\to \sin^22x=1$
$\to \sin2x=1\to 2x=\dfrac{\pi}{2}+k2\pi\to x=\dfrac{\pi}{4} +k\pi$
$\to \sin2x=-1\to 2x=\dfrac32\pi+k2\pi\to x=\dfrac34\pi+k\pi$
