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VysotskaDi
@VysotskaDi
July 2022
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Ребят,срочно,помогите решить 1,3,4 задание
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Dимасuk
Verified answer
1. cos²3x - sin²3x = 0
-cos²6x = 0
cos6x = 0
6x = π/2 + πn, n€Z.
x = π/12 + πn/6, n€Z.
Ответ: х = π/12 + πn, n€Z.
3. 5sin²x + sinx - 6 = 0
Пусть t = sinx, t€[-1; 1]
5t² + t - 6 = 0
D = 1 + 6•4•5 = 121 = 11²
t1 = (-1 + 11)/10 = 1
t2 = (-1 - 11)/10 = -12/10 - не уд. условию
Обратная замена:
sinx= 1
x = π/2 + 2πn, n€Z.
Ответ: х = π/2 + 2πn, n€Z.
4. 1 - cos2x = 2sinx
1 - (1 - 2sin²x) = 2sinx
2sin²x = 2sinx
sinx = 1
x = π/2 + 2πn, n€Z.
sinx = 0
x = πn, n€Z.
Ответ: х = πn, n€Z; π/2 + 2πn, n€Z.
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Answers & Comments
Verified answer
1. cos²3x - sin²3x = 0-cos²6x = 0
cos6x = 0
6x = π/2 + πn, n€Z.
x = π/12 + πn/6, n€Z.
Ответ: х = π/12 + πn, n€Z.
3. 5sin²x + sinx - 6 = 0
Пусть t = sinx, t€[-1; 1]
5t² + t - 6 = 0
D = 1 + 6•4•5 = 121 = 11²
t1 = (-1 + 11)/10 = 1
t2 = (-1 - 11)/10 = -12/10 - не уд. условию
Обратная замена:
sinx= 1
x = π/2 + 2πn, n€Z.
Ответ: х = π/2 + 2πn, n€Z.
4. 1 - cos2x = 2sinx
1 - (1 - 2sin²x) = 2sinx
2sin²x = 2sinx
sinx = 1
x = π/2 + 2πn, n€Z.
sinx = 0
x = πn, n€Z.
Ответ: х = πn, n€Z; π/2 + 2πn, n€Z.