2 câu trả lời
Đáp án:
$ x = -\dfrac π4 + kπ$, $k\in\mathbb Z$
Lời giải:
$\sin x + \cos x = 0$
$\Leftrightarrow\dfrac 1{\sqrt2}\sin x +\dfrac 1{\sqrt2}\cos x = 0$
$\Leftrightarrow\cos \dfrac π4.\sin x +\sin \dfrac π4.\cos x = 0$
$\Leftrightarrow\sin\left({x +\dfrac π4}\right) = 0$
$\Leftrightarrow x +\dfrac π4 = kπ$
$\Leftrightarrow x = -\dfrac π4 + kπ$, $k\in\mathbb Z$
Vậy $ x = -\dfrac π4 + kπ$, $k\in\mathbb Z$
Đáp án:
$S=\left\{-\dfrac{\pi}{4}+k\pi\,\bigg{|}\,k\in\mathbb Z\right\}$
Giải thích các bước giải:
$\sin x+\cos x=0$
$⇒\dfrac{\sqrt 2}{2}.\left(\sin x+\cos x\right)=0$
$⇒\dfrac{\sqrt 2}{2}.\sin x+\dfrac{\sqrt 2}{2}.\cos x=0$
$⇒\sin x.\cos \dfrac{\pi}{4}+\cos x.\sin\dfrac{\pi}{4}=0$
$⇒\sin\left(x+\dfrac{\pi}{4}\right)=0$
$⇒x+\dfrac{\pi}{4}=k\pi\,\,(k\in\mathbb Z)$
$⇒x=-\dfrac{\pi}{4}+k\pi\,\,(k\in\mathbb Z)$
Vậy $S=\left\{-\dfrac{\pi}{4}+k\pi\,\bigg{|}\,k\in\mathbb Z\right\}$.