1.пусть cos x=a<span>2a2-3a+1=0</span>D=9-8=1<span>a1=1/2, a2=1</span>cosx=1/2 или cosx=1<span>x=±arccos 1/2+2Пn x=2Пk </span>x=±П/3+2Пn2.<span>1+7cos2x-6sinxcosx=0 ,(sin2x=2sinxcosx)</span><span>sin2x+8cos2x-6sinxcosx=0 ,(1=sin2x+cos2x) разделим на cos2x≠0</span><span>tg2x-6tgx+8=0</span>D=4tgx=2 или tgx=4<span><span>x=arctg2+Пn </span> <span> x=arctg4+Пk</span></span>