a.
Đặt \(sinx+cosx=t\) với \(\left|t\right|\le\sqrt{2}\)
\(\Rightarrow2sinx.cosx=t^2-1\)
Pt trở thành:
\(2t=2\left(t^2-1\right)+1\)
\(\Leftrightarrow2t^2-2t-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{1+\sqrt{3}}{2}\\t=\dfrac{1-\sqrt{3}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1+\sqrt{3}}{2}\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1-\sqrt{3}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{5\pi}{12}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{7\pi}{12}+k2\pi\\x+\dfrac{\pi}{4}=-\dfrac{\pi}{12}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{3}+k2\pi\\x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
d.
Đặt \(sinx-cosx=t\) với \(\left|t\right|\le\sqrt{2}\)
\(\Rightarrow sin2x=1-t^2\)
pt trở thành:
\(2\left(1-t^2\right)=t-1\)
\(\Leftrightarrow2t^2+t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
