Toán Giải phương trình: $\frac{cos^3x-cos^2x}{sinx+cosx}$=2(1+sinx) 11/08/2021 By Aaliyah Giải phương trình: $\frac{cos^3x-cos^2x}{sinx+cosx}$=2(1+sinx)
Giải thích các bước giải: $\dfrac{\cos^3x-\cos^2x}{\sin x+\cos x}=2(1+\sin x)$ $\to \dfrac{\cos^2x(\cos x-1)}{\sin x+\cos x}=2(1+\sin x)$ $\to \dfrac{(1-\sin^2x)(\cos x-1)}{\sin x+\cos x}=2(1+\sin x)$ $\to \dfrac{(1-\sin x)(1+\sin x)(\cos x-1)}{\sin x+\cos x}=2(1+\sin x)$ $\to 1+\sin x=0\to x=\dfrac{3\pi}{2}+k2\pi$ Hoặc $\dfrac{(1-\sin x)(\cos x-1)}{\sin x+\cos x}=2$ $\to(1-\sin x)(\cos x-1)=2(\sin x+\cos x)$ $\to\sin x+\cos x+\cos x.\sin x+1=0$ $\to (\sin x+1)(\cos x+1)=0$ $\to \cos x+1=0\to x=\pi+k2\pi$ Trả lời
Giải thích các bước giải:
$\dfrac{\cos^3x-\cos^2x}{\sin x+\cos x}=2(1+\sin x)$
$\to \dfrac{\cos^2x(\cos x-1)}{\sin x+\cos x}=2(1+\sin x)$
$\to \dfrac{(1-\sin^2x)(\cos x-1)}{\sin x+\cos x}=2(1+\sin x)$
$\to \dfrac{(1-\sin x)(1+\sin x)(\cos x-1)}{\sin x+\cos x}=2(1+\sin x)$
$\to 1+\sin x=0\to x=\dfrac{3\pi}{2}+k2\pi$
Hoặc $\dfrac{(1-\sin x)(\cos x-1)}{\sin x+\cos x}=2$
$\to(1-\sin x)(\cos x-1)=2(\sin x+\cos x)$
$\to\sin x+\cos x+\cos x.\sin x+1=0$
$\to (\sin x+1)(\cos x+1)=0$
$\to \cos x+1=0\to x=\pi+k2\pi$