Question

Giúp mình với ạ

Answer 1

\(\Leftrightarrow\dfrac{\sqrt{2}}{2}sinx+\dfrac{\sqrt{2}}{2}cosx=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

Answer 2

c)    \(sinx+cosx=1\)

   \(\Rightarrow\)\(\sqrt{2}sin\left(x+\dfrac{\pi}{2}\right)=1\) \(\Rightarrow sin\left(x+\dfrac{\pi}{2}\right)=\dfrac{1}{\sqrt{2}}\)\(=sin\dfrac{\pi}{4}\)

   \(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{\pi}{2}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{2}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

   \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)