Đáp án: $ \left\{\begin{array}{l} x=\dfrac{-\pi}{12}+k\dfrac{\pi}{2}\\ x=\dfrac{\pi}{4}+k\dfrac{\pi}{2}\end{array} \right.(k\in\mathbb Z)$
Giải thích các bước giải:
Ta có: ${\sin}^4x+{\cos}^4x=\left({{\sin}^2x+{\cos}^2x}\right)^2-2{\sin}^2x{\cos}^2x$
$=1-2\left({\dfrac{\sin2x}{2}}\right)^2$
$=1-\dfrac{{\sin}^22x}{2}$ $=1-\dfrac{1}{2}\dfrac{1-\cos 4x}{2}$
$=\dfrac{3}{4}+\dfrac{\cos 4x}{4}$
Thay vào phương trình ta được:
$4\left({\dfrac{3}{4}+\dfrac{\cos 4x}{4}}\right)+\sqrt3\sin4x=2$
$\Rightarrow 3+\cos 4x+\sqrt3\sin4x=2$
$\Rightarrow \cos 4x+\sqrt3\sin4x=-1$
$\Rightarrow\dfrac{1}{2}\cos 4x+\dfrac{\sqrt3}{2}\sin4x=\dfrac{-1}{2}$
$\Rightarrow\sin \left({4x+\dfrac{\pi}{6}}\right)=\dfrac{-1}{2}$
$\Rightarrow \left[\begin{array}{l} 4x+\dfrac{\pi}{6}=\dfrac{-\pi}{6}+k2\pi\\ 4x+\dfrac{\pi}{6}=\dfrac{7\pi}{6}+k2\pi\end{array} \right.(k\in\mathbb Z)$
$\Rightarrow \left[\begin{array}{l} x=\dfrac{-\pi}{12}+k\dfrac{\pi}{2}\\ x=\dfrac{\pi}{4}+k\dfrac{\pi}{2}\end{array} \right.(k\in\mathbb Z)$
