Решить уравнение 25*sin(x)cos(x)-sin(x)-cos(x)=5

Task/25521524
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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

$25\sin x\cos x-\sin x-\cos x= 5\\ 25\sin x\cos x-(\sin x+\cos x)=5(\sin^2x+\cos ^2x)\\ 25\sin x\cos x-(\sin x+\cos x)=5((\sin x+\cos x)^2-2\sin x \cos x)\\ 25\sin x\cos x-(\sin x+\cos x)=5(\sin x+\cos x)^2-10\sin x\cos x\\ 5(\sin x+\cos x)^2+\sin x+\cos x-35\sin x\cos x=0$
Введём замену $\sin x+\cos x=t$ , тогда $1+2\sin x\cos x=t^2;\\ \sin x\cos x= \frac{t^2-1}{2}$ , имеем

$5t^2+t- \frac{35(t^2-1)}{2} =0\\ \\ 10t^2+2t-35t^2+35=0\\ \\ 25t^2-2t-35=0$

$t_1= \frac{1-2 \sqrt{219} }{25}$
$t_2=\frac{1+2 \sqrt{219} }{25}$

Возвращаемся к обратной замене
$\sin x+\cos x=\frac{1\pm2 \sqrt{219} }{25} \\ \\ \sqrt{2} \sin(x+ \frac{\pi}{4})=\frac{1\pm2 \sqrt{219} }{25} \\ \\ \sin (x+ \frac{\pi}{4}) = \frac{1\pm2 \sqrt{219} }{25 \sqrt{2} } \\ \\ \\ x=(-1)^k\arcsin(\frac{1\pm2 \sqrt{219} }{25 \sqrt{2} }) - \frac{\pi}{4}+ \pi k,k \in Z$