c.
Đặt \(sinx-cosx=t\) với \(\left|t\right|\le\sqrt{2}\)
\(\Leftrightarrow sin^2x+cos^2x-sin2x=t^2\)
\(\Rightarrow sin2x=1-t^2\)
Pt trở thành:
\(1-t^2=5\left(t-1\right)\)
\(\Leftrightarrow t^2+5t-6=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-6\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow sinx-cosx=1\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
f.
Đặt \(sinx+cosx=t\) với \(\left|t\right|\le\sqrt{2}\)
\(\Rightarrow sin2x=t^2-1\)
Pt trở thành:
\(t=t^2-1\)
\(\Leftrightarrow t^2-t-1=0\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{1+\sqrt{5}}{2}>\sqrt{2}\left(loại\right)\\t=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1-\sqrt{5}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=arcsin\left(\dfrac{1-\sqrt{5}}{2\sqrt{2}}\right)+k2\pi\\x+\dfrac{\pi}{4}=\pi-arcsin\left(\dfrac{1-\sqrt{5}}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arcsin\left(\dfrac{1-\sqrt{5}}{2\sqrt{2}}\right)+k2\pi\\x=\dfrac{3\pi}{4}-arcsin\left(\dfrac{1-\sqrt{5}}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
