2 câu trả lời
Đáp án:
$\left\{ \begin{array}{l} x = k2\pi \\ x = \dfrac{\pi }{2} + k2\pi \end{array} \right.(k\in\mathbb Z)$
Giải thích các bước giải:
$\begin{array}{l} \sin x + \cos x = 1 \Leftrightarrow \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 1\\ \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }} = \sin \dfrac{\pi }{4}\\ \Leftrightarrow \left[ \begin{array}{l} x + \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\ x + \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = k2\pi \\ x = \dfrac{\pi }{2} + k2\pi \end{array} \right. (k\in\mathbb Z)\end{array}$
Vậy phương trình có nghiệm:
$\left\{ \begin{array}{l} x = k2\pi \\ x = \dfrac{\pi }{2} + k2\pi \end{array} \right.(k\in\mathbb Z)$.
Đáp án:
$S=\left\{k2\pi;\dfrac{\pi}{2}+k2\pi\,\bigg{|}\,k\in\mathbb Z\right\}$
Giải thích các bước giải:
$\sin x+\cos x=1$
$⇔\dfrac{\sqrt 2}{2}.\sin x+\dfrac{\sqrt 2}{2}.\cos x=\dfrac{\sqrt 2}{2}$
$⇔\sin x.\cos\dfrac{\pi}{4}+\cos x.\sin\dfrac{\pi}{4}=\dfrac{\sqrt 2}{2}$
$⇔\sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt 2}{2}$
$⇔\left[ \begin{array}{l}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$
$⇔\left[ \begin{array}{l}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$
Vậy $S=\left\{k2\pi;\dfrac{\pi}{2}+k2\pi\,\bigg{|}\,k\in\mathbb Z\right\}$.
