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appel1997
@appel1997
July 2022
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zarembo73
A) cos²x-cosx-2=0;
cosx=t; -1≤t≤1;
t²-t-2=0;
D=1+8=9;
t1=(1-3)/2=-2/2=-1;
t2=(1+3)/2=4/2=2.
cosx=-1;
x=π+2πn, n∈Z.
Ответ: π+2πn, n∈Z.
б) 3cos²x-2sinx+2=0;
3(1-sin²x)-2sinx+2=0;
3-3sin²x-2sinx+2=0;
-3sin²x-2sinx+5=0;
3sin²x+2sinx-5=0;
sinx=t, -1≤t≤1;
3t²+2t-5=0;
D=4+60=64;
t1=(-2-8)/6=-10/6=-5/3;
t2=(-2+8)/6=6/6=1.
sinx=1;
x=π/2+2πn, n∈Z.
Ответ: π/2+2πn, n∈Z.
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Answers & Comments
cosx=t; -1≤t≤1;
t²-t-2=0;
D=1+8=9;
t1=(1-3)/2=-2/2=-1;
t2=(1+3)/2=4/2=2.
cosx=-1;
x=π+2πn, n∈Z.
Ответ: π+2πn, n∈Z.
б) 3cos²x-2sinx+2=0;
3(1-sin²x)-2sinx+2=0;
3-3sin²x-2sinx+2=0;
-3sin²x-2sinx+5=0;
3sin²x+2sinx-5=0;
sinx=t, -1≤t≤1;
3t²+2t-5=0;
D=4+60=64;
t1=(-2-8)/6=-10/6=-5/3;
t2=(-2+8)/6=6/6=1.
sinx=1;
x=π/2+2πn, n∈Z.
Ответ: π/2+2πn, n∈Z.