a.
Đặt \(sinx+cosx=t\in\left[-\sqrt{2};\sqrt{2}\right]\)
\(\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)
Phương trình trở thành:
\(2t+t^2-1+1=0\)
\(\Rightarrow t\left(t+2\right)=0\Rightarrow\left[{}\begin{matrix}t=0\\t=-2< -\sqrt{2}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow sinx+cosx=0\)
\(\Rightarrow tanx=-1\)
\(\Rightarrow x=-\dfrac{\pi}{4}+k\pi\)
a, Đặt \(sinx+cosx=t\left(t\in\left[-\sqrt{2};\sqrt{2}\right]\right)\)
\(pt\Leftrightarrow2t+t^2-1+1=0\)
\(\Leftrightarrow t^2+2t=0\)
\(\Leftrightarrow t\left(t+2\right)=0\)
\(\Leftrightarrow t=0\)
\(\Leftrightarrow sinx+cosx=0\)
\(\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\)
b.
Đặt \(sinx-cosx=t\in\left[-\sqrt{2};\sqrt{2}\right]\)
\(\Rightarrow sinx.cosx=\dfrac{1-t^2}{2}\)
Phương trình trở thành:
\(2\sqrt{2}t-2\left(1-t^2\right)=1\)
\(\Leftrightarrow2t^2+2\sqrt{2}t-3=0\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{\sqrt{2}}{2}\\t=-\dfrac{3\sqrt{2}}{2}< -\sqrt{2}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)
\(\Rightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Rightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{5}+k2\pi\end{matrix}\right.\)
\(\Rightarrow...\)
b, Đặt \(sinx-cosx=t\left(t\in\left[-\sqrt{2};\sqrt{2}\right]\right)\)
\(pt\Leftrightarrow2\sqrt{2}\left(sinx-cosx\right)-2sin2x=1\)
\(\Leftrightarrow2\sqrt{2}t+2\left(t^2-1\right)=1\)
\(\Leftrightarrow2t^2+2\sqrt{2}t-3=0\)
\(\Leftrightarrow t=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow x+\dfrac{\pi}{4}=\pm\dfrac{\pi}{3}+k2\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k2\pi\\x=-\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\)
