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1sin(x-π/3)=-1x-π/3=-π/2+2πnx=-π/6+2πn,n∈z2tg(π/6-x)=-1/√3tg(x-π/6)=1/√3x-π/6=π/6+πnx=π/3+πn,n∈z3cosx=a2a²+5a+2=0D=25-16=9a1=(-5-3)/4=-2⇒cosx=-2<-1 нет решенияa2=(-5+3)/4=-1/2⇒cosx=-1/2⇒x=+-2π/3+2πn,n∈z42cos2xcosx=0 U sinx≠-1⇒x≠-π/2+2πn,n∈zcos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2,n∈zcosx=0⇒x=π/2+πk,k∈z с учетом ОДЗ x=π/2+2πk,k∈z5sinx>-√3/2x∈(-π/3+2πn;4π/3+2πn,n∈z)6tg(π/3-x)≤1/√3tg(x-π/3)≥-1/√3-π/6+πn≤x-π/3<π/2+πnπ/6+πn≤x<5π/6+πnx∈[π/6+πn;5π/6+πn,n∈z)