1.
ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\)
\(1+tanx=2\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow1+\dfrac{sinx}{cosx}=2\left(sinx+cosx\right)\)
\(\Leftrightarrow\dfrac{sinx+cosx}{cosx}=2\left(sinx+cosx\right)\)
\(\Leftrightarrow\left(2-\dfrac{1}{cosx}\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\dfrac{\left(2cosx-1\right)\left(sinx+cosx\right)}{cosx}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{1}{2}\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{4}=k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x=-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)
4.
\(sin3x+cos3x-sinx+cosx=\sqrt{2}cos2x\)
\(\Leftrightarrow2cos2x.sinx+2cos2x.cosx=\sqrt{2}cos2x\)
\(\Leftrightarrow cos2x.\left(2sinx+2cosx-\sqrt{2}\right)=0\)
\(\Leftrightarrow cos2x.\left[\sqrt{2}\left(sinx+cosx\right)-1\right]=0\)
\(\Leftrightarrow cos2x.\left[2sin\left(x+\dfrac{\pi}{4}\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{2}\end{matrix}\right.\)
Đến đây dễ rồi tự làm tiếp.
2.
\(cos2x+\sqrt{3}sin2x+\sqrt{3}sinx-cosx=2\)
\(\Leftrightarrow\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x+\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=1\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)+sin\left(x-\dfrac{\pi}{6}\right)=1\)
\(\Leftrightarrow1-2sin^2\left(x-\dfrac{\pi}{6}\right)+sin\left(x-\dfrac{\pi}{6}\right)=1\)
\(\Leftrightarrow2sin^2\left(x-\dfrac{\pi}{6}\right)-sin\left(x-\dfrac{\pi}{6}\right)=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)\left[2sin\left(x-\dfrac{\pi}{6}\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{6}\right)=0\\sin\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=k\pi\\x-\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=\dfrac{\pi}{3}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
3.
\(\sqrt{3}cos5x-2sin3x.cos2x-sinx=0\)
\(\Leftrightarrow\sqrt{3}cos5x-2.\dfrac{1}{2}\left(sin5x+sinx\right)-sinx=0\)
\(\Leftrightarrow\sqrt{3}cos5x-sin5x-2sinx=0\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cos5x-\dfrac{1}{2}sin5x=sinx\)
\(\Leftrightarrow cos\left(5x+\dfrac{\pi}{6}\right)=sinx\)
\(\Leftrightarrow cos\left(5x+\dfrac{\pi}{6}\right)=cos\left(\dfrac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+\dfrac{\pi}{6}=\dfrac{\pi}{2}-x+k2\pi\\5x+\dfrac{\pi}{6}=-\dfrac{\pi}{2}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}6x=\dfrac{\pi}{3}+k2\pi\\4x=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{18}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\end{matrix}\right.\)
