`sin^2 x+ sin^2 2x = 0`
`<=> (1-cos2x) + (1-cos4x) =0`
<=> cos2x + cos4x = 2`
`<=> cos2x + 2cos^2 2x - 1 - 2 =0`
`<=> 2cos^2 2x + cos2x - 3 =0`
`<=>` \(\left[ \begin{array}{l}cos2x=1\\cos2x=\dfrac{-3}{2} (L)\end{array} \right.\)
`<=> 2x = k2π`
`<=> x =kπ`
Vậy `x= kπ (k \in \mathbbZ)`
.
`cos 2x + cos 4x = 2`
`<=> cos2x + 2cos^2 2x - 1 - 2=0`
`<=> 2cos^2 2x + cos2x - 3 =0`
`<=>....`
`<=> x=kπ `
