Ответ:
x = πk, k∈Z
Объяснение:
x =\pi k" alt="5cos^2x-sin^2x=5\\/sin^2x = 1-cos^2x/\\5cos^2x -1+cos^2x=5\\cos^2x=1\\\\\left \{ {{cosx=1} \atop {cosx=-1}} \right. \left \{ {{x=2\pi k} \atop {x=\pi+2\pi k}} \right. => x =\pi k" align="absmiddle" class="latex-formula">