Ответы на вопрос:
1)cosx< 0⇒x∈(π/2+2πn; 3π/2+2πn,n∈z) -cosx+√3sinx=0 2(√3/2sinx-1/2cosx)=0 2sin(x-π/6)=0 x-π/6=πn x=π/6+πn u x∈(π/2+2πn; 3π/2+2πn,n∈z)⇒x=7π/6+2πn 2π≤7π/6+2πn≤7π/2 12≤7+12n≤21 5≤12n≤14 5/12≤n≤7/6 n=1⇒x=7π/6+2π=19π/6 2)cosx≥0⇒x∈[-π/2+2πk; π/2+2πk,k∈z] cosx+√3sinx=0 2sin(x+π/6)=0 x+π/6=πk x=-π/6+πk u x∈[-π/2+2πk; π/2+2πk,k∈z]⇒x=π/6+2πk 2π≤π/6+2πk≤7π/2 12≤1+12k≤21 11≤12k≤20 11/12≤k≤5/3 k=1⇒x=π/6+2π=13π/6
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