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batalygina
@batalygina
July 2022
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12cos^2x+sinx-11=0
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smartman3
12(1-sin^2x) + sinx - 11 = 0.
Замена sinx->t
12-12t^2+t-11=0
12t^2-t-1=0
D=49
T1= (1+7)/24=1/3
T2=-1/4
Sinx = 1/3, x= arcsin 1/3
X2=arcsin-1/4
2 votes
Thanks 1
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Answers & Comments
Замена sinx->t
12-12t^2+t-11=0
12t^2-t-1=0
D=49
T1= (1+7)/24=1/3
T2=-1/4
Sinx = 1/3, x= arcsin 1/3
X2=arcsin-1/4