1.
\(\left(sin2x+\sqrt{3}cos2x\right)^2=2cos\left(2x-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left(\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x\right)^2=\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow sin^2\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow cos^2\left(\dfrac{\pi}{2}-2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow cos^2\left(2x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{6}\right)\left[cos\left(2x-\dfrac{\pi}{6}\right)-\dfrac{1}{2}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(2x-\dfrac{\pi}{6}\right)=0\\cos\left(2x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\\2x-\dfrac{\pi}{6}=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}+k\pi\\x=-\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)
2.
ĐK: \(x\ne\dfrac{k\pi}{2};x\ne\pi+k2\pi\)
\(cotx+\left(1+tanx.tan\dfrac{x}{2}\right).sinx=4\)
\(\Leftrightarrow\dfrac{cosx}{sinx}+\left(1+\dfrac{sinx.sin\dfrac{x}{2}}{cosx.cos\dfrac{x}{2}}\right).sinx=4\)
\(\Leftrightarrow\dfrac{cosx}{sinx}+\dfrac{cosx.cos\dfrac{x}{2}+sinx.sin\dfrac{x}{2}}{cosx.cos\dfrac{x}{2}}.sinx=4\)
\(\Leftrightarrow\dfrac{cosx}{sinx}+\dfrac{cos\left(x-\dfrac{x}{2}\right).sinx}{cosx.cos\dfrac{x}{2}}=4\)
\(\Leftrightarrow\dfrac{cosx}{sinx}+\dfrac{cos\dfrac{x}{2}.sinx}{cosx.cos\dfrac{x}{2}}=4\)
\(\Leftrightarrow\dfrac{cosx}{sinx}+\dfrac{sinx}{cosx}=4\)
\(\Leftrightarrow\dfrac{cos^2x+sin^2x}{sinx.cosx}=4\)
\(\Leftrightarrow\dfrac{2}{sin2x}=4\)
\(\Leftrightarrow sin2x=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{6}+k2\pi\\2x=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)
3.
ĐK: \(x\ne\dfrac{k\pi}{2}\)
\(cotx-tanx=\dfrac{2}{sin2x}-4sin2x\)
\(\Leftrightarrow\dfrac{cos^2x-sin^2x}{sinx.cosx}=\dfrac{2}{sin2x}-4sin2x\)
\(\Leftrightarrow\dfrac{2cos2x}{sin2x}=\dfrac{2-4sin^22x}{sin2x}\)
\(\Leftrightarrow\dfrac{2cos2x}{sin2x}=\dfrac{2cos2x}{sin2x}\)
\(\Leftrightarrow x\ne\dfrac{k\pi}{2}\)
