4.
a, \(sinx-cosx=1\)
\(\Leftrightarrow sin^2x+cos^2x-2sinx.cosx=1\)
\(\Leftrightarrow sinx.cosx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{k\pi}{2}\)
Thử lại vào phương trình đã cho ta được \(x=\dfrac{\pi}{2}+k2\pi;x=\pi+k2\pi\)
d, \(sin2x+cos2x=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow2x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{8}+k\pi\)
5.
\(\left|sinx+\dfrac{1}{2}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow sin^2x+sinx+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow sinx\left(sinx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
