Question

Решить уравнение. применив подстановку y = cosx - sinx, решите уравнение: 4-4(cosx - sinx) = sin2x

Answer 1

$4-4(\cos x-\sin x)=\sin2x\\ 4(\sin^2x+\cos^2x)-4(\cos x-\sin x)=\sin2x\\ 4(\sin^2x+\cos^2x-\sin2x+\sin2x)-4(\cos x-\sin x)=\sin 2x\\ 4(\cos x-\sin x)^2+4\sin2x-4(\cos x-\sin x)=\sin 2x\\ 4(\cos x-\sin x)^2-4(\cos x-\sin x)+3\sin2x=0$
Пусть $\cos x-\sin x=t\,(|t| \leq \sqrt{2} )$, тогда возведем обе части до квадрата и получаем 1-sin2x=t², откуда sin2x=1-t²
$4t^2-4t+3(1-t^2)=0\\ 4t^2-4t+3-3t^2=0\\ t^2-4t+3=0$
По т. Виета
t1 = 1
t2 = 3 - не удовлетворяет условию при |t|√2

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

Answer 2

$4-4(cosx-sinx)=sin2x\\\\y=cosx-sinx\\y^2=(cosx-sinx)^2\\y^2=cos^2x-2sinx*cosx+sin^2x\\y^2=1-2sinx*cosx\\y^2-1=-2sinx*cosx\\\\4-4(cosx-sinx)-sin2x=0\\4-4(cosx-sinx)-2sinx*cosx=0\\4-4y+y^2-1=0\\y^2-4y+3=0\\D=16-12=4\\\\y_1=\frac{4+2}{2}=\frac{6}2=3\\\\y_2=\frac{4-2}2=\frac{2}2=1$

Обратная подстановка.
$y=cosx-sinx\\\\ \left[\begin{array}{ccc}cosx-sinx=1\\cosx-sinx=3\end{array}\right=\ \textgreater \ \left[\begin{array}{ccc}-\sqrt2sin(x-\frac{\pi}4)=1\\-\sqrt2sin(x-\frac{\pi}4)=3\end{array}\right=\ \textgreater \ \\\left[\begin{array}{ccc}sin(x-\frac{\pi}4)=\frac{\sqrt2}2\\sin(x-\frac{\pi}4) \neq -\frac{3\sqrt2}2\end{array}\right=\ \textgreater \ sinx(x-\frac{\pi}4) =\frac{\sqrt2}2\\\\ \left[\begin{array}{ccc}x-\frac{\pi}4=\frac{\pi}4+2\pi n;n\in Z\\x-\frac{\pi}4=\frac{3\pi}4+2\pi n;n\in Z\end{array}\right=\ \textgreater \$ $\left[\begin{array}{ccc}x=\frac{\pi}2+2\pi n;n\in Z\\x=\pi+2\pi n;n\in Z\end{array}\right$