степень внизу) (4х+3) 1,2 - id253444720220624 от angelinakuznetsova 24.06.2022 03:58

Question

Алгебра: степень внизу) (4х+3) 1,2

angelinakuznetsova · Accepted Answer

Task/25521524

Решить уравнение
25*sin(x)cos(x)-sin(x)-cos(x)=5 ;
25*( ( sin(x) +cos(x) )² - 1) /2 -  ( sin(x) +cos(x) =5 ;
замена:  t = sin(x) +cos(x) = √2cos(x -π/4) ; -√2 ≤ √2cos(x -π/4)  ≤  √2
25(t² -1)/2 - t =5 ;
25t² -2t -35 =0 ; D₁ =(2/2)² - 25*(-35) =1 +875 =876 =(2√219)²
t₁ = (1 -2√219) / 25 ;
t₂ =  (1+2√219) / 25 .
* * * t₁ и t₂ ∈ [ - √2 ; √2] * * *
a)
√2cos(x -π/4) = (1 -2√219) / 25  ;
cos(x -π/4) = √2(1 -2√219) / 50
x -π/4 = ± arccos (√2(1 -2√219) /  50) +2πn , n ∈ Z .
x = π/ 4 ± arccos (√2(1 -2√219) /  50) +2πn , n ∈ Z .
б)
√2cos(x -π/4) = (1 +2√219) / 25;
x = π/ 4 ± arccos (√2(1 +2√219) / 50) +2πn , n ∈ Z .√2

степень внизу) (4х+3) > 1,2

MOGZ ответил