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ЯЯЯNNN
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August 2022
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cos2x+8sinx=3 Помогите пожалуйста
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irka1804
Cos2x = 1 - 2sin^2x
1 - 2sin^2x + 8sinx = 3
2sin^2x - 8sinx +2 = 0
sinx = t
t^2 - 4t + 1 = 0
t = (4 +- √(16 - 4))/2 = (4 +- 2√3)/2 = 2 +- √3
Подходит t = 2 - √3
sinx = 2 - √3
x = arcsin(2 - √3) + 2πn, n ∈ Z и x = π - arcsin(2-√3) + 2πn
, n ∈ Z
2 votes
Thanks 2
shevtsovavalia
Cos2x+8sinx=3
cos²x-sin²x+8sinx-3sin²x-3cos²x=0
-2cos²x-4sin²x+8sinx=0
-2+2sin²x-4sin²x+8sinx=0
2sin²x-8sinx+2=0
sinx=2-√2
x=(-1)ⁿarcsin(2-√2)+πn
0 votes
Thanks 3
ЯЯЯNNN
я не поняла Вы по какой то формуле решили?
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Answers & Comments
1 - 2sin^2x + 8sinx = 3
2sin^2x - 8sinx +2 = 0
sinx = t
t^2 - 4t + 1 = 0
t = (4 +- √(16 - 4))/2 = (4 +- 2√3)/2 = 2 +- √3
Подходит t = 2 - √3
sinx = 2 - √3
x = arcsin(2 - √3) + 2πn, n ∈ Z и x = π - arcsin(2-√3) + 2πn, n ∈ Z
cos²x-sin²x+8sinx-3sin²x-3cos²x=0
-2cos²x-4sin²x+8sinx=0
-2+2sin²x-4sin²x+8sinx=0
2sin²x-8sinx+2=0
sinx=2-√2
x=(-1)ⁿarcsin(2-√2)+πn