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pavelkulinarny
@pavelkulinarny
August 2022
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12cosx - 5 sinx = -6, 5
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Dимасuk
Verified answer
12cosx - 5sinx = -6,5
Введём дополнительный угол С:
C = √12² + 5² = 13
Разделим на 13:
cosx·12/13 - 5/13sinx = -1/2
cosx·cos(arccos(12/13)) - 5sin(arccos(12/13)sinx = -1/2
cos(x + arccos(12/13)) = -1/2
x + arccos(12/13) = ±2π/3 + 2πn, n ∈ Z
x = ±2π/3 - arccos(12/13) + 2πn, n ∈ Z
2 votes
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sedinalana
Cosx=(1-tg²x/2))/(1+tg²x/2),sinx=2tgx/2/(1+tg²x/2)
tgx/2=u⇒cosx=(1-u²)/(1+u²) ,sinx=2u/(1+u²)
-------------------------------------------
12cosx-5sinx=-13/2
24cosx-10sinx=-13
24*(1-u²)/(1+u²)-10*2u/(1+u²)=-13
24(1-u²)-20u=-13(1+u²)=0
24-24u²-20u+13+13u²=0
11u²-20u-37=0
D=400+1628=2028
√D=26√3
u1=(20-26√3)/22=(10-13√3)/11 U u2=(10+13√3)/11
tgx/2=(10-13√3)/11⇒x/2=arctg[(10-13√3)/11]+πk⇒
x=2arctg[(10-13√3)/11]+πk,k∈z
tgx/2=(10+13√3)/11⇒x/2=arctg[(10+13√3)/11]+πk⇒
x=2arctg[(10+13√3)/11]+πk,k∈z
0 votes
Thanks 0
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Answers & Comments
Verified answer
12cosx - 5sinx = -6,5Введём дополнительный угол С:
C = √12² + 5² = 13
Разделим на 13:
cosx·12/13 - 5/13sinx = -1/2
cosx·cos(arccos(12/13)) - 5sin(arccos(12/13)sinx = -1/2
cos(x + arccos(12/13)) = -1/2
x + arccos(12/13) = ±2π/3 + 2πn, n ∈ Z
x = ±2π/3 - arccos(12/13) + 2πn, n ∈ Z
tgx/2=u⇒cosx=(1-u²)/(1+u²) ,sinx=2u/(1+u²)
-------------------------------------------
12cosx-5sinx=-13/2
24cosx-10sinx=-13
24*(1-u²)/(1+u²)-10*2u/(1+u²)=-13
24(1-u²)-20u=-13(1+u²)=0
24-24u²-20u+13+13u²=0
11u²-20u-37=0
D=400+1628=2028
√D=26√3
u1=(20-26√3)/22=(10-13√3)/11 U u2=(10+13√3)/11
tgx/2=(10-13√3)/11⇒x/2=arctg[(10-13√3)/11]+πk⇒
x=2arctg[(10-13√3)/11]+πk,k∈z
tgx/2=(10+13√3)/11⇒x/2=arctg[(10+13√3)/11]+πk⇒
x=2arctg[(10+13√3)/11]+πk,k∈z