e.
Đặt \(sinx+cosx=t\) với \(\left|t\right|\le\sqrt{2}\)
\(\Rightarrow sin2x=t^2-1\)
Pt trở thành:
\(3t=2\left(t^2-1\right)\)
\(\Leftrightarrow2t^2-3t-2=0\)
\(\Rightarrow\left[{}\begin{matrix}t=2\left(loại\right)\\t=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow sinx+cosx=-\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=arcsin\left(-\dfrac{\sqrt{2}}{4}\right)+k2\pi\\x+\dfrac{\pi}{4}=\pi-arcsin\left(-\dfrac{\sqrt{2}}{4}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arcsin\left(-\dfrac{\sqrt{2}}{4}\right)+k2\pi\\x=\dfrac{3\pi}{4}-arcsin\left(-\dfrac{\sqrt{2}}{4}\right)+k2\pi\end{matrix}\right.\)
b.
Đặt \(sinx+cosx=t\) (với \(\left|t\right|\le\sqrt{2}\))
\(\Rightarrow sin^2x+cos^2x+2sinx.cosx=t^2\)
\(\Rightarrow sin2x=t^2-1\)
Pt trở thành:
\(t^2-1-3t+3=0\)
\(\Leftrightarrow t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2>\sqrt{2}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow sinx+cosx=1\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
