Ответы на вопрос:
12sinπ/4*cos(π/4-x)=√2 2*√2/2*sin(x+π/4)=√2 sin(x+π/4)=1 x+π/4=π/2+2πn x=-π/4+π/2+2πn x=π/4+2πn,n∈z 2 2sinx+tgx*ctgx=0 2sinx+1=0 2sinx=-1 sinx=-1/2 x=-5π/6+2πn,n∈z -π< -5π/6+2πn< π -6< -5+12n< 6 -1< 12n< 11 -1/12< n< 11/12 n=0⇒x=-5π/6 x=-π/6+2πn -π< -π/6+2π< π -6< -1+12n< 6 -5< 12n< 7 -5/12< n< 7/12 n=0πx=-π/6 -5π/6+(-π/6)=-π
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