2sin^22x=1-cos4x;\\1-cos4x-5cos4x=1-6cos4x=0;\\cos4x=\frac{1}{6} ; 4x=бarccos\frac{1}{6} +2pi*n" alt="cos4x=1-2sin^22x => 2sin^22x=1-cos4x;\\1-cos4x-5cos4x=1-6cos4x=0;\\cos4x=\frac{1}{6} ; 4x=бarccos\frac{1}{6} +2pi*n" align="absmiddle" class="latex-formula">
Ответ: x=±1/4*arccos(1/6)+pi*n/2, n∈Z.