giải như sau:
sin^3x.sin3x + cos^3x.cos3x= căn2/4
\Leftrightarrow (3sinx-sin3x)/4.sin3x+\{3cosx-cos3x}/4.cos3x= căn2/4
\Leftrightarrow( 3sinx-sin3x)sin3x+(3cosx-cos3x)cos3x= căn2
\Leftrightarrow 3sinx.sin3x-sin^2(3x)+3cosxcos3x-cos^2(3x)= căn 2
\Leftrightarrow 3sinxsin3x+3cosxcos3x-(sin^2(3x)+cos^2(3x))= căn 2
\Leftrightarrow 3sinxsin3x+3cosxcos3x=căn 2+1
\Leftrightarrow 3/2(cos2x-cos4x)+3/2(cos2x+cos4x)=căn 2+1
\Leftrightarrow 3/2(cos2x-cos4x+cos2x+cos4x)= căn 2 +1
\Leftrightarrow 3/2.2cos2x=căn 2 +1
\Leftrightarrow cos2x={căn2+1}/3
\Leftrightarrow 2x =+_arccos{căn2+1}/3+k2pj
\Leftrightarrow x=+_(arccos\frac{căn2+1}/3)/2+kpj