Giúp mik vs mik đang cần gấp, mik cảm ơn

`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)`

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`cos 2x + cos 4x = 2`

`<=> cos2x + 2cos^2 2x – 1 – 2=0`

`<=> 2cos^2 2x + cos2x – 3 =0`

`<=>….`

`<=> x=kπ `