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emmafall
@emmafall
July 2022
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Уравнение sin2x+√3*sinx=0
Найти корни на промежутке [5pi/2;7pi/2]
если решение уравнения:
x=пk, k ∈ Z
x= + - 5п/6 + 2пk, k ∈ Z
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volnovs
Verified answer
2 sinx cosx +
√3sinx=0
sinx (2cosx+
√3)=0
1)sin x=0
x=Пn
2)2cosx+
√3=0
cosx=-
√3/2
x=+-5П/6+2Пn
ОТБОР КОРНЕЙ
x=3п-п/6=17п/6
х=5п/2+п/2=3п
х=3п+п/6=19п/6
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Answers & Comments
Verified answer
2 sinx cosx + √3sinx=0sinx (2cosx+√3)=0
1)sin x=0
x=Пn
2)2cosx+√3=0
cosx=-√3/2
x=+-5П/6+2Пn
ОТБОР КОРНЕЙ
x=3п-п/6=17п/6
х=5п/2+п/2=3п
х=3п+п/6=19п/6