Giải phương trình: sin2x – (sinx + cosx – 1)(2sinx – cosx – 3) = 0 01/10/2021 Bởi Maya Giải phương trình: sin2x – (sinx + cosx – 1)(2sinx – cosx – 3) = 0
Đáp án: $x=\dfrac12k\pi$ Giải thích các bước giải: Ta có: $\sin2x-(\sin x+\cos x-1)(2\sin x-\cos x-3)=0$ $\to \sin2x=(\sin x+\cos x-1)(2\sin x-\cos x-3)$ $\to 2\sin x\cos x=(\sin x+\cos x-1)(2\sin x-\cos x-3)$ $\to (2\sin x\cos x+1)-1=(\sin x+\cos x-1)(2\sin x-\cos x-3)$ $\to (2\sin x\cos x+\sin^2x+\cos^2x)-1=(\sin x+\cos x-1)(2\sin x-\cos x-3)$ $\to (\sin x+\cos x)^2-1=(\sin x+\cos x-1)(2\sin x-\cos x-3)$ $\to (\sin x+\cos x-1)(\sin x+\cos x+1)=(\sin x+\cos x-1)(2\sin x-\cos x-3)$ $\to (\sin x+\cos x-1)(\sin x+\cos x+1)-(\sin x+\cos x-1)(2\sin x-\cos x-3)=0$ $\to (\sin x+\cos x-1)(\sin x+\cos x+1-2\sin x+\cos x+3)=0$ $\to (\sin x+\cos x-1)(2\cos x-\sin x+4)=0$ Mà $2\cos x-\sin x+4\ge -2-1+4>0$ $\to \sin x+\cos x-1=0$ $\to \sin x+\cos x=1$ $\to (\sin x+\cos x)^2=1$ $\to 1+\sin2x=1$ $\to \sin2x=0$ $\to 2x=k\pi$ $\to x=\dfrac12k\pi$ Bình luận
Đáp án: $x=\dfrac12k\pi$
Giải thích các bước giải:
Ta có:
$\sin2x-(\sin x+\cos x-1)(2\sin x-\cos x-3)=0$
$\to \sin2x=(\sin x+\cos x-1)(2\sin x-\cos x-3)$
$\to 2\sin x\cos x=(\sin x+\cos x-1)(2\sin x-\cos x-3)$
$\to (2\sin x\cos x+1)-1=(\sin x+\cos x-1)(2\sin x-\cos x-3)$
$\to (2\sin x\cos x+\sin^2x+\cos^2x)-1=(\sin x+\cos x-1)(2\sin x-\cos x-3)$
$\to (\sin x+\cos x)^2-1=(\sin x+\cos x-1)(2\sin x-\cos x-3)$
$\to (\sin x+\cos x-1)(\sin x+\cos x+1)=(\sin x+\cos x-1)(2\sin x-\cos x-3)$
$\to (\sin x+\cos x-1)(\sin x+\cos x+1)-(\sin x+\cos x-1)(2\sin x-\cos x-3)=0$
$\to (\sin x+\cos x-1)(\sin x+\cos x+1-2\sin x+\cos x+3)=0$
$\to (\sin x+\cos x-1)(2\cos x-\sin x+4)=0$
Mà $2\cos x-\sin x+4\ge -2-1+4>0$
$\to \sin x+\cos x-1=0$
$\to \sin x+\cos x=1$
$\to (\sin x+\cos x)^2=1$
$\to 1+\sin2x=1$
$\to \sin2x=0$
$\to 2x=k\pi$
$\to x=\dfrac12k\pi$