4 × (sin^4x +cos^4x) + √3 × sin4x = 2

4(sin^4 x + cos^4 x) + √3sin4x = 2

<=> 4[(sin^2 x + cos^2 x)^2 - 2sin^2 x . cos^2 x] + √3sin4x = 2

<=> 4[1 - (1/2)sin^2 2x] + √3sin4x = 2

<=> -2sin^2 2x + √3sin4x = -2

<=> cos4x - 1 + √3sin4x = 2

<=> cos4x + √3sin4x = -1

<=> cos4x + tanpi/3.sin4x = -1

<=> cos4x.cospi/3 + sinpi/3.sin4x = -cospi/3

<=> cos(4x - pi/3) = cos(pi - pi/3) = cos(2pi/3)

=>

[4x - pi/3 = 2pi/3 + k2pi

[4x - pi/3 = -2pi/3 + k2pi

<=>

[x = pi/4 + kpi/2

[x = -pi/12 + kpi/2

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:

$4\left({\dfrac{3}{4}+\dfrac{\cos 4x}{4}}\right)+\sqrt3\sin4x=2$

$\Rightarrow 3+\cos 4x+\sqrt3\sin4x=2$

$\Rightarrow \cos 4x+\sqrt3\sin4x=-1$

$\Rightarrow\dfrac{1}{2}\cos 4x+\dfrac{\sqrt3}{2}\sin4x=\dfrac{-1}{2}$

$\Rightarrow\sin \left({4x+\dfrac{\pi}{6}}\right)=\dfrac{-1}{2}$

$\Rightarrow \left[\begin{array}{l} 4x+\dfrac{\pi}{6}=\dfrac{-\pi}{6}+k2\pi\\ 4x+\dfrac{\pi}{6}=\dfrac{7\pi}{6}+k2\pi\end{array} \right.(k\in\mathbb Z)$

$\Rightarrow \left[\begin{array}{l} x=\dfrac{-\pi}{12}+k\dfrac{\pi}{2}\\ x=\dfrac{\pi}{4}+k\dfrac{\pi}{2}\end{array} \right.(k\in\mathbb Z)$