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larisakravcenko

1)  a) ОДЗ:  2cosx + √3 ≠ 0; sinx > 0 ⇒ cosx ≠  - 2/√3 ⇒ x ≠ ± ( π - π/6) + 2πk, k ∈ Z;  x ≠ 5/6 π +2πk, k ∈ Z;   ;  b) log₂(sinx) · (  log₂(sinx)  + 1) = 0 ⇒ 1.  log₂(sinx) = 0 ⇒ sinx = 2⁰ = 1 ⇒ x = π/2 +2πn, n ∈ Z    2. log₂(sinx)  = - 1;   sinx =  2⁻¹ = 1/2 ⇒ x₁ = π/6 + 2πl; l ∈ Z;   X₂ = 5/6 π + 2πm; m ∈ Z  - посторонний, так как ∉ ОДЗ ⇒ ОТВЕТ:  x₁ = π/6 + 2πl; l ∈ Z;  x = π/2 +2πn, n ∈ Z ;    2) ОДЗ: sinx > 0;  Замена: = t = log₈(sinx);   3t² - 5t - 2 = 0;  t = 2;  - 1/3 ⇒  log₈(sinx) = 2;  sinx = 8² = 16; так как |sinxI ≤ 1 - данное уравнение не имеет решений;   log₈(sinx) = - 1/3;  sinx = 8⁻¹/³ = 1/2 ⇒ sinx = 1/2 ⇒ x= π/6 + 2πn, n ∈ Z;  x = 5/6 π + 2πk, k ∈ Z;  b) π/6; 5/6 π ∈ [ 0; 3/2 π];  Ответ: a) x= π/6 + 2πn, n ∈ Z; x = 5/6 π + 2πk, k ∈ Z; b) x= π/6;               x = 5/6 π  

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