4) \(sinx=\dfrac{1}{\sqrt{3}}\left(0< x< \dfrac{\pi}{2}\right)\)
\(sin^2x+cos^2x=1\Rightarrow cos^2x=1-sin^2x=1-\dfrac{1}{3}=\dfrac{2}{3}\Rightarrow cosx=\dfrac{\sqrt{2}}{\sqrt{3}}\left(0< x< \dfrac{\pi}{2}\right)\)
\(cos2x=cos^2x-sin^2x=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
\(cos\left(x+\dfrac{\pi}{3}\right)=cosx.cos\dfrac{\pi}{3}-sinx.sin\dfrac{\pi}{3}=\dfrac{\sqrt{2}}{\sqrt{3}}.\dfrac{1}{2}-\dfrac{1}{\sqrt{3}}.\dfrac{\sqrt[]{3}}{2}=\dfrac{1}{\sqrt{6}}-\dfrac{1}{2}=\dfrac{2-\sqrt{6}}{2\sqrt{6}}\)
\(tanx=\dfrac{sinx}{cosx}=\dfrac{\dfrac{1}{\sqrt{3}}}{\dfrac{\sqrt{2}}{\sqrt{3}}}=\dfrac{1}{\sqrt[]{2}}\)
\(tan\left(x+\dfrac{\pi}{3}\right)=\dfrac{tanx+tan\dfrac{\pi}{3}}{1+tanx.tan\dfrac{\pi}{3}}=\dfrac{\dfrac{1}{\sqrt{2}}+\sqrt{3}}{1+\dfrac{1}{\sqrt{2}}\sqrt{3}}=\dfrac{\dfrac{1+\sqrt{6}}{\sqrt{2}}}{\dfrac{1+\sqrt{3}}{\sqrt{2}}}=\dfrac{1+\sqrt{6}}{1+\sqrt{3}}\)
5) a) \(cos\left(4x-\dfrac{2\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(4x-\dfrac{2\pi}{3}\right)=cos\dfrac{\pi}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{2\pi}{3}=\dfrac{\pi}{3}+k2\pi\\4x-\dfrac{2\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\pi+k2\pi\\4x=\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\dfrac{\pi}{2}\\x=\dfrac{\pi}{12}+k\dfrac{\pi}{2}\end{matrix}\right.\)
b) \(cos2x+\sqrt{3}.sin2x=0\)
\(\Leftrightarrow\dfrac{1}{2}.cos2x+\dfrac{\sqrt{3}}{2}.sin2x=0\)
\(\Leftrightarrow cos\dfrac{\pi}{3}.cos2x+sin\dfrac{\pi}{3}.sin2x=0\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow2x-\dfrac{\pi}{3}=\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\dfrac{5\pi}{12}+k\dfrac{\pi}{2}\left(k\in Z\right)\)
