<span>cos2x+2tgx=2
Сos2x = 2 - 2tgx
Cos</span>²x - Sin²x = 2(1 - Sinx/Cosx)
(Cosx - Sinx)(Cosx + Sinx) = 2(Cosx -Sinx)/Cosx
(Cosx - Sinx)(Cosx + Sinx) - 2(Cosx -Sinx)/Cosx = 0
(Cosx - Sinx)(Cosx +Sinx -2/Cosx) = 0
(Cosx - Sinx) = 0| : Сosx или (Cosx +Sinx -2/Cosx) = 0
1 - tgx = 0 Cos²x +SinxCosx -2 = 0
tgx = 1 Cos²x + SinxCosx -2*1 = 0
x = π/4 + πk , k∈Z Cos²x + SinxCosx -2(Sin²x + Cos²x) =0
Cos²x + SinxCosx -2Sin²x -2Cos²x =0
-Cos²x + SinxCosx -2Sin²x =0 | :Сos²x
-1 +tgx -2tg²x = 0
tgx = t
2t² -t +1 = 0
D< 0
∅
Х²+у²-8х+2у+17=0
(х²-8х+16)+(у²+2у+1)=0
(х-4)²+(у+1)²=0
(х-4)²>0;(у+1)²>0
х-4=0;х=4
у+1=0;у=-1
ответ х=4;у=-1
Решение в фотке:
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