Đáp án:
`\(\left[ \begin{array}{l}x=\dfrac{\pi}{3}+k\pi\\x=-\dfrac{\pi}{3}+k\pi\end{array} \right.\)`(k inZZ)`
Giải thích các bước giải:
`8cos^4x=1+cos4x`
`<=>8.(cos^2x)^2=1+cos4x`
`<=>8.((1+cos2x)/2)^2=1+cos4x`
`<=>2.(1+cos2x)^2=1+cos4x`
`<=>2(1+2cos2x+cos^2 2x)=1+2cos^2 2x-1`
`<=>2+4cos2x=0`
`<=>cos2x=-1/2`
`<=>cos2x=cos((2pi)/3)`
`<=>`\(\left[ \begin{array}{l}2x=\dfrac{2\pi}{3}+k2\pi\\2x=-\dfrac{2\pi}{3}+k2\pi\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}{3}+k\pi\\x=-\dfrac{\pi}{3}+k\pi\end{array} \right.\)`(k inZZ)`
