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lera5590
@lera5590
August 2022
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решить уравнение:
2cos^2x= 3 sinx
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nKrynka
Решение
2*(1 - sin∧2x) - 3sinx = 0
2sin∧2x + 3sinx - 2 = 0
d + 9 + 4*2*2 = 25
sinx = (-3 - 5) /4 = -2 - не удовлетворяет ОДЗ: I sinx I ≤1
sinx = (-3 + 5)/4 = 1/2
x = (-1)∧n arcsin(1/2) + πn, n∈Z
x = (-1)∧n (π/6) + πn, n∈Z
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Answers & Comments
2*(1 - sin∧2x) - 3sinx = 0
2sin∧2x + 3sinx - 2 = 0
d + 9 + 4*2*2 = 25
sinx = (-3 - 5) /4 = -2 - не удовлетворяет ОДЗ: I sinx I ≤1
sinx = (-3 + 5)/4 = 1/2
x = (-1)∧n arcsin(1/2) + πn, n∈Z
x = (-1)∧n (π/6) + πn, n∈Z