Câu 1 Sin 5x - cos5x+1=0 Câu 2 sin3x +cos3x =1

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
\sin 5x - \cos 5x + 1 = 0\\
 \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin 5x - \frac{{\sqrt 2 }}{2}\cos 5x =  - \frac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \sin 5x.\cos \frac{\pi }{4} - \cos 5x.\sin \frac{\pi }{4} = \sin \left( {\frac{{ - \pi }}{4}} \right)\\
 \Leftrightarrow \sin \left( {5x - \frac{\pi }{4}} \right) = \sin \left( { - \frac{\pi }{4}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
5x - \frac{\pi }{4} =  - \frac{\pi }{4} + k2\pi \\
5x - \frac{\pi }{4} = \frac{{5\pi }}{4} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{k2\pi }}{5}\\
x = \frac{{3\pi }}{{10}} + \frac{{k2\pi }}{5}
\end{array} \right.\\
2,\\
\sin 3x - \cos 3x = 1\\
 \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin 3x - \frac{{\sqrt 2 }}{2}\cos 3x = \frac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \sin 3x.\cos \frac{\pi }{4} - \cos 3x.\sin \frac{\pi }{4} = \sin \left( {\frac{\pi }{4}} \right)\\
 \Leftrightarrow \sin \left( {3x - \frac{\pi }{4}} \right) = \sin \left( {\frac{\pi }{4}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
3x - \frac{\pi }{4} = \frac{\pi }{4} + k2\pi \\
3x - \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{k2\pi }}{3}\\
x = \frac{\pi }{3} + \frac{{k2\pi }}{3}
\end{array} \right.
\end{array}\)