ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\)
\(4sin^2x+tanx+\sqrt{2}\left(1+tanx\right)sin3x=1\)
\(\Leftrightarrow4sin^2x-2+1+tanx+\sqrt{2}\left(1+tanx\right)sin3x=0\)
\(\Leftrightarrow-2cos2x+\left(1+tanx\right)\left(1+\sqrt{2}sin3x\right)=0\)
\(\Leftrightarrow2sin^2x-2cos^2x+\dfrac{sinx+cosx}{cosx}\left(1+\sqrt{2}sin3x\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2sinx-2cosx+\dfrac{1+\sqrt{2}sin3x}{cosx}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\2sinx-2cosx+\dfrac{1+\sqrt{2}sin3x}{cosx}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\2sinx.cosx-2cos^2x+1+\sqrt{2}sin3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\sin2x-cos2x+\sqrt{2}sin3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)+\sqrt{2}sin3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\2sin\left(\dfrac{5x}{2}-\dfrac{\pi}{8}\right).cos\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\sin\left(\dfrac{5x}{2}-\dfrac{\pi}{8}\right)=0\\cos\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=k\pi\\\dfrac{5x}{2}-\dfrac{\pi}{8}=k\pi\\\dfrac{x}{2}+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
Pt tương đương
\(\left\{{}\begin{matrix}4sin^2x+\dfrac{sinx}{cosx}+\sqrt{2}\left(1+\dfrac{sinx}{cosx}\right).sin3x=1\\cosx\ne0\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}4sin^2x.cosx+sinx+\sqrt{2}\left(sinx+cosx\right).sin3x=cosx\\cosx\ne0\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}2sin2x.sinx+sinx+\sqrt{2}\left(sinx+cosx\right).sin3x=cosx\\cosx\ne0\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}sinx-cos3x+\sqrt{2}\left(sinx+cosx\right).sin3x=0\\cosx\ne0\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}sinx-sin\left(3x+\dfrac{\pi}{2}\right)-2sin\left(x+\dfrac{\pi}{4}\right).sin3x=0\\cosx\ne0\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}2cos\left(2x+\dfrac{\pi}{4}\right).sin\left(-x-\dfrac{\pi}{4}\right)-2sin\left(x+\dfrac{\pi}{4}\right).sin3x=0\\cosx\ne0\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}2cos\left(2x+\dfrac{\pi}{4}\right).sin\left(x+\dfrac{\pi}{4}\right)+2sin\left(x+\dfrac{\pi}{4}\right).sin3x=0\\cosx\ne0\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)\left[cos\left(2x+\dfrac{\pi}{4}\right)+sin3x\right]=0\\cosx\ne0\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\sin3x=sin\left(2x+\dfrac{3\pi}{4}\right)\end{matrix}\right.\\x\ne\dfrac{\pi}{2}+k\pi,k\in Z\end{matrix}\right.\)
Giải nốt nhé, toàn phương trình cơ bản rồi
