mọi người giúp e vs sin^6x+cos^6x=7/16

Đáp án:

$\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k\pi\\x=\dfrac{\pi}{3}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\\x=\dfrac{2\pi}{3}+k\pi\end{array} \right.$

Giải thích các bước giải:

 $sin^6x+cos^6x=\dfrac{7}{16}$

$(sin^2x)^3+(cos^2x)^3=\dfrac{7}{16}$

$\Rightarrow (sin^2x+cos^2x)(sin^4x-sin^2x.cos^2x+cos^4x)=\dfrac{7}{16}$

$\Rightarrow (sin^2x+cos^2x)((sin^2x+cos^2x)^2-3sin^2x.cos^2x)=\dfrac{7}{16}$

$\Rightarrow 1.(1-\dfrac{3}{4}sin^2 2x)=\dfrac{7}{16}$

$\Rightarrow sin^2 2x=\dfrac{3}{4}$

$\Rightarrow \left[ \begin{array}{l}sin2x=\dfrac{\sqrt{3}}{2}\\sin2x=-\dfrac{\sqrt{3}}{2}\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}2x=\dfrac{\pi}{3}+k2\pi\\2x=\dfrac{2\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}+k2\pi\\2x=\dfrac{4\pi}{3}+k2\pi\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}x=\dfrac{\pi}{6}+k\pi\\x=\dfrac{\pi}{3}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\\x=\dfrac{2\pi}{3}+k\pi\end{array} \right.$