Đáp án:
$x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$
Giải thích các bước giải:
$sin^4x+cos^4x=2(sin^6x+cos^6x)+\dfrac{5}{4}cos2x$
$\rightarrow (sin^2x+cos^2x)^2-2sin^2xcos^2x=2((sin^2x+cos^2x)^3-3sin^2xcos^2x(sin^2x+cos^2x))+\dfrac{5}{4}cos2x$
$\rightarrow 1-2sin^2xcos^2x=2(1-3sin^2xcos^2x)+\dfrac{5}{4}cos2x$
$\rightarrow 1-4sin^2xcos^2x+\dfrac{5}{4}cos2x=0$
$\rightarrow 1-sin^22x+\dfrac{5}{4}cos2x=0$
$\rightarrow cos^22x+\dfrac{5}{4}cos2x=0$
$\rightarrow cos2x(cos2x+\dfrac{5}{4})=0$
$\rightarrow cos2x=0$
$\rightarrow 2x=\dfrac{\pi}{2}+k\pi$
$\rightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$
