sin2x-2sinx-2cosx+2=0

Đáp án:

$S=\left\{k2\pi;\dfrac{\pi}{2}+k2\pi\,\bigg{|}\,k\in\mathbb Z\right\}$

Giải thích các bước giải:

$\sin2x-2\sin x-2\cos x+2=0$

$⇔2\sin x\cos x-2\sin x-2\cos x+2=0$

$⇔\sin x\cos x-(\sin x+\cos x)+1=0$

Đặt $t=\sin x+\cos x=\sqrt 2\sin\left(x+\dfrac{\pi}{4}\right)\,(t\in[-\sqrt 2;\sqrt 2])$

$⇒t^2=\sin^2x+\cos^2x+2\sin x\cos x⇒\sin x\cos x=\dfrac{t^2-1}{2}$

$\text{Pt}⇔\dfrac{t^2-1}{2}-t+1=0$

$⇔t^2-1-2t+2=0$

$⇔t^2-2t+1=0$

$⇔(t-1)^2=0$

$⇔t=1\,(TM)$

$⇔\sqrt 2\sin\left(x+\dfrac{\pi}{4}\right)=1$

$⇔\sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt 2}{2}$

$⇔\left[ \begin{array}{l}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$

$⇔\left[ \begin{array}{l}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$

Vậy $S=\left\{k2\pi;\dfrac{\pi}{2}+k2\pi\,\bigg{|}\,k\in\mathbb Z\right\}$.