(2сos^2 2x-5cosx+2)*√ sinx = 0 1) 2cos^2 2x-5cosx+2=0 2(сos^2 x-sin^2 x)-5cosx+2=0 2(cos^2 x-(1-cos^2 x)-5cosx+2=0 2(cos^2 x+cos^2 x-1)-5cosx+2=0 4cos^2 x-5cosx+2-2=0 4cos^2-5cosx=0 cделаем замену cosx=t 4t^2-5t=0 t(4t-5)=0 t=0 cosx=0 4t-5=0 4t=5 t=5/4 =1,25 cosx=1,25 неверно, так как -1≤cos≤1