Найти корни sinx+1/2=0 x∈[0; 3p] sinx-1/2=0 x∈[-p/2; 3p/2] sinx+√2/2=0 x∈[-3p; 0] sinx-√2/2=0 x∈[-3p/2; 5p/2] sinx+√3/2=0 x∈[-2p; 2p]

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zombdanidani

Алгебра

4,4(61 оценок)

1) sinx = -1/2; x = (-1)^(n+1)* arcsin(|-1/2|) + pi*n; x = (-1)^(n+1)* pi/6) + pi*n; n ∈ z n = 0; x = -pi/6 ∉[0; 3p] n = 1; x = pi/6 + pi = 7pi/6 ∈ [0; 3p]n = 2; x = -pi/6 + 2pi = 11pi/6 ∈ [0; 3p]n = 3; x = pi/6 + 3pi ∉ [0; 3p] ответ: x = 7pi/6 ∪ x = 11pi/6 2) sinx = 1/2; x = (-1)^(n)* arcsin1/2) + pi*n; x = (-1)^(n)* pi/6)+ pi*n; n ∈ z n = -1; x = -pi/6 - pi ∉ [-p/2; 3p/2] n = 0; x = pi/6 ∈[-p/2; 3p/2] n = 1; x = -pi/6 + pi = 5pi/6 ∈[-p/2; 3p/2] n = 2; x = pi/6 + 2pi ∉[-p/2; 3p/2] ответ: x = pi/6 ∪ x = 5pi/6 3) sinx = -√2/2; x = (-1)^(n+1)* arcsin(|-√2/2|) + pi*n; x = (-1)^(n+1)* pi/4) + pi*n; n ∈ z n = -4; x = -pi/4 - 4pi ∉[-3p; 0] n = -3; x = pi/4 - 3pi = -11pi/4 ∈[-3p; 0] n = -2; x = -pi/4 -2pi = -9pi/4 ∈[-3p; 0] n = -1; x = pi/4 - pi = - 3pi/4 ∈[-3p; 0] n = 0; x = -pi/4 ∈[-3p; 0] n = 1; x = pi/4 + pi ∉[-3p; 0] ответ: x = -11pi/4 ∪ x = -9pi/4 ∪ x = pi/4 - pi ∪ x = -pi/4 4) sinx = √2/2; x = (-1)^(n)* arcsin(√2/2) + pi*n; x = (-1)^(n)* pi/4)+ pi*n; n ∈ z n = -2; x = pi/4 - 2pi = -7pi/4 ∉[-3p/2; 5p/2] n = -1; x = -pi/4 - pi = - 5pi/4 ∈[-3p/2; 5p/2] n = 0; x = pi/4 ∈[-3p/2; 5p/2] n = 1; x = -pi/4 + pi = 3pi/4 ∈ [-3p/2; 5p/2]n = 2; x = pi/4 + 2pi = 9pi/4 ∈ [-3p/2; 5p/2]n = 3; x = -pi/4 + 3pi ∉[-3p/2; 5p/2] ответ: x = -5pi/4 ∪ x = pi/4 ∪ x = 3pi/4 ∪ x = 9pi/4 5) sinx = -√3/2; x = (-1)^(n+1)* arcsin(|-√3/2|) + pi*n; x = (-1)^(n+1)* pi/3) + pi*n; n ∈ z n = -2; x = -pi/3 - 2pi ∉[-2p; 2p] n = -1; x = pi/3 - pi = -2pi/3; n = 0; x = -pi/3 ∈[-2p; 2p] n = 1; x = pi/3 + pi = 4pi/3 ∈[-2p; 2p] n = 2; x = -pi/3 + 2pi = 5pi/3 ∈[-2p; 2p] n = 3; x = pi/3 + 3pi ∉[-2p; 2p] ответ: x = -2pi/3 ∪ x = -pi/3 ∪ x =4pi/3 ∪ x = 5pi/3