Giải thích các bước giải:
Ta có:
$sin^2(4x)-sin^2(6x)=0\\⇔\frac{1-cos8x}{2}-\frac{1-cos12x}{2}=0\\⇔1-cos8x-1+cos12x=0\\⇔cos12x=cos8x\\⇔\left \{ {{12x=8x+k2\pi} \atop {12x=-8x+k2\pi}} \right.\\⇔\left \{ {{4x=k2\pi} \atop {20x=k2\pi}} \right.\\⇔ \left \{ {{x=\frac{k \pi}{2}} \atop {x=\frac{k \pi}{10}}} \right. (k∈Z) $
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