Có: `sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2x cos^2x`
`=1-1/2 .sin^2 2x`
`=1-1/2 .(1-cos4x)/2`
`=1/4 cos4x+3/4`
`->` PT: `4(1/4 cos 4x+3/4)+\sqrt3 sin4x=2`
`<=> cos4x + \sqrt3 sin4x = -1`
`<=> \sqrt3/2 sin4x + 1/2 cos4x=-1/2`
`<=> sin(4x+π/6)=-1/2`
`<=>` \(\left[\begin{array}{l} 4x+\dfrac{π}{6}=\dfrac{-π}{6}+k2π\\ 4x+\dfrac{π}{6}=\dfrac{7π}{6}+k2\pi\end{array} \right.\)
`<=>` \(\left[\begin{array}{l} x=\dfrac{-π}{12}+\dfrac{kπ}{2}\\ x=\dfrac{π}{4}+\dfrac{kπ}{2}\end{array} \right. (k \in \mathbb Z)\)
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