Найдите корни уравнения tgx+1=0, принажлежащие отрезку [0;2п]
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tqx+1 = 0 ; x∈ [ 0;2π ].
tqx = -1 ;
x = -π/4 +π*k , k∈Z .
x₁ = -π/4 +π =3π/4 , при k =1;
x₂ = - π/4 +2π =7π/4 , при k =2.
ответ: 3π/4 ; 7π/4 .
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0 ≤-π/4 +π*k ≤2π;
π/4 ≤π*k ≤ π/4+2π;
1/4 ≤ k ≤ 9/4; k∈Z ⇔ 1≤ k ≤2. ⇒ k =1;2