Đáp án:
`sin^2x = cos^2\2x +cos^2\3x `
`<=>{1-cos2x}/2={1+cos4x}/2={1+cos6x}/2`
`<=>1-cos2x=1+cos4x+1+cos6x`
`<=>cos6x+cos4x+cos2x+1=0`
`<=>2cdotcos5xcdotcosx+2cos^2x=0`
`<=>2cosx(cos5x+cosx)=0`
`<=>4cosxcdotcos3xcdotcos2x=0`
`<=>`\(\left[ \begin{array}{l}\cos x=0\\\cos 2x=0\\\cos 3x =0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{6}+\dfrac{k\pi}{5}\end{array} \right.\) `(k in mathbbZ)`
Vì `x=(0;pi)`
`=>` Có các `n_0:x={\frac{\pi}{6};\frac{\pi}{4};\frac{3\pi}{2};{5pi}/6;pi/2}`
`->` Tổng các `n_0={5pi}/2`
