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wolfstarrkk
@wolfstarrkk
July 2022
1
15
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Cos²x=1/2
и
Sin²x=1
прошу решить с объяснением, тк я не была на этой теме
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nKrynka
Verified answer
1) cos^2x = 1/2
cosx = - √2/2
x1 = (+ -)arccos(-√2/2) + 2πn, n∈Z
x1 = (+ -)(π - π/4) + 2πn, n∈z
x1 = (+ -)(3π/4) + 2πn, n∈Z
2) cosx = √2/2
x2 = (+ -)arccos(√2/2) + 2πk, n∈Z
x2 = (+ -)(π/4) + 2πk, n∈Z
2) sin^2x = 1
a) sinx = -1
x1 = - π/2 + 2πm, m∈Z
b) sinx = 1
x2 = π/2 + 2πk, k∈Z
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Answers & Comments
Verified answer
1) cos^2x = 1/2cosx = - √2/2
x1 = (+ -)arccos(-√2/2) + 2πn, n∈Z
x1 = (+ -)(π - π/4) + 2πn, n∈z
x1 = (+ -)(3π/4) + 2πn, n∈Z
2) cosx = √2/2
x2 = (+ -)arccos(√2/2) + 2πk, n∈Z
x2 = (+ -)(π/4) + 2πk, n∈Z
2) sin^2x = 1
a) sinx = -1
x1 = - π/2 + 2πm, m∈Z
b) sinx = 1
x2 = π/2 + 2πk, k∈Z