mn giúp eGiải pt sinx+cosx=1

Đáp án:

Giải thích các bước giải: \[\begin{array}{l} \sin x + \cos x = 1\\ \Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = 1\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \frac{1}{{\sqrt 2 }} = \sin \frac{\pi }{4}\\ \Leftrightarrow \left[ \begin{array}{l} x + \frac{\pi }{4} = \frac{\pi }{4} + k2\pi \\ x + \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = k2\pi \\ x = \frac{\pi }{2} + k2\pi \end{array} \right. \end{array}\]

\[\begin{array}{l}
sinx + cosx = 1\\
 \Leftrightarrow \sqrt 2 sin(x + \frac{\pi }{4}) = 1\\
 \Leftrightarrow sin(x + \frac{\pi }{4}) = 1\sqrt 2  = sin\frac{\pi }{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \frac{\pi }{4} = \frac{\pi }{4} + k2\pi \\
x + \frac{\pi }{4} = \frac{{3\pi }}{4}k2\pi 
\end{array} \right.\,\,\, \Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \frac{\pi }{2} + k2\pi 
\end{array} \right.
\end{array}\]