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ironwoman9
@ironwoman9
July 2022
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найдите число корней уравнения tg6x*sin12x+cos12x-cos24x=0 на промежутке [0;2п]
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kirichekov
Verified answer
Tg6x=sin6x/cos6x
sin12x=2sin6x*cos6x
cos12x=cos²6x-sin²6x
tg6x*sin12x+cos12x=(sin6x/cos6x)*(2sin6x*cos6x)+cos²6x-sin²6x=2sin²6x+cos²6x-sin²6x=sin²6x+cos²6x=1
1-cos24x=0, cos24x=1. ( частный случай: cost=1, t=2πn, n∈Z)
24x=2πn, n∈Z
x=(πn)/12, n∈Z
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Answers & Comments
Verified answer
Tg6x=sin6x/cos6xsin12x=2sin6x*cos6x
cos12x=cos²6x-sin²6x
tg6x*sin12x+cos12x=(sin6x/cos6x)*(2sin6x*cos6x)+cos²6x-sin²6x=2sin²6x+cos²6x-sin²6x=sin²6x+cos²6x=1
1-cos24x=0, cos24x=1. ( частный случай: cost=1, t=2πn, n∈Z)
24x=2πn, n∈Z
x=(πn)/12, n∈Z