Đáp án: $\sin x\in\{0,1\}$
Giải thích các bước giải:
Ta có:
$(1+\sqrt5)(\sin x-\cos x)+\sin2x-1-\sqrt5=0$
$\to (1+\sqrt5)(\sin x-\cos x)+2\sin x\cos x-(\sin^2x+\cos^2x)-\sqrt5=0$
$\to (1+\sqrt5)(\sin x-\cos x)-(\sin^2x-2\sin x\cos x+\cos^2x)-\sqrt5=0$
$\to (1+\sqrt5)(\sin x-\cos x)-(\sin x-\cos x)^2-\sqrt5=0$
Đặt $\sin x-\cos x=t$
$\to (1+\sqrt5)t-t^2-\sqrt5=0$
$\to t\in\{1,\sqrt5\}$
Mà $t=\sin x-\cos x$
$\to t\cdot \dfrac1{\sqrt2}=\sin(x-\dfrac14\pi)\le 1$
$\to t\le\sqrt2$
$\to t=1$
$\to \sin x-\cos x=1$
$\to \sin x-1=\cos x$
$\to (\sin x-1)^2=\cos^2x$
$\to\sin^2x-2\sin x+1=1-\sin^2x$
$\to 2\sin^2x-2\sin x=0$
$\to 2\sin x(\sin x-1)=0$
$\to \sin x\in\{0,1\}$
