Đáp án:
`cos^{2}x+cos^{2}2x+cos^{2}3x+cos^{2}4x=2`
`<=>(1)/(2).(cos2x+1)+(1)/(2).(cos4x+1)+(1)/(2).(cos6x+1)+(1)/(2).(cos8x+1)=2`
`<=>(1)/(2).(cos2x+1+cos4x+1+cos6x+1+cos8x+1)=2`
`<=>cos2x+cos4x+cos6x+cos8x+4=4`
`<=>cos2x+cos4x+cos6x+cos8x=0`
`<=>(cos2x+cos8x)+(cos4x+cos6x)=0`
`<=>2.cos((8x-2x)/(2)).cos((8x+2x)/(2))+2.cos((6x-4x)/(2)).cos((6x+4x)/(2))=0`
`<=>2cos3x.cos5x+2cosx.cos5x=0`
`<=>2.cos5x.(cos3x+cosx)=0`
`<=>2.cos5x.2.cosx.cos2x=0`
`<=>4.cosx.cos2x.cos5x=0`
`<=>`\(\left[ \begin{array}{l}cosx=0\\cox2x=0\\cox5x=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=\dfrac{π}{2}+kπ\\2x=\dfrac{π}{2}+kπ\\5x=\dfrac{π}{2}+kπ\end{array}(k∈Z) \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{π}{2}+kπ\\x=\dfrac{π}{4}+\dfrac{kπ}{2}\\x=\dfrac{π}{10}+\dfrac{kπ}{5}\end{array}(k∈Z) \right.\)
