4.
\(0< x< \dfrac{\pi}{2}\Rightarrow\dfrac{\pi}{6}< x+\dfrac{\pi}{6}< \dfrac{2\pi}{3}\)
Mà \(sin\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}\Rightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}\)
\(\Rightarrow x=\dfrac{\pi}{6}\)
\(\Rightarrow cos\left(x+\dfrac{5\pi}{12}\right)=cos\left(\dfrac{\pi}{6}+\dfrac{5\pi}{12}\right)=cos\left(\dfrac{7\pi}{12}\right)\) (tới đây bấm máy hoặc làm tay như sau):
\(=cos\left(\dfrac{\pi}{3}+\dfrac{\pi}{4}\right)=cos\left(\dfrac{\pi}{3}\right).cos\left(\dfrac{\pi}{4}\right)-sin\left(\dfrac{\pi}{3}\right).sin\left(\dfrac{\pi}{4}\right)=\dfrac{1}{2}.\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{3}}{2}.\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}-\sqrt{6}}{4}\)
5.
\(\dfrac{cosx-3cos2x+cos3x}{sinx-3sin2x+sin3x}=\dfrac{cosx+cos3x-3cos2x}{sinx+sin3x-3sin2x}\)
\(=\dfrac{2cosx.cos2x-3cos2x}{2sin2x.cosx-3sin2x}=\dfrac{cos2x\left(2cosx-3\right)}{sin2x\left(2cosx-3\right)}=\dfrac{cos2x}{sin2x}=cot2x=\dfrac{1}{tan2x}\)
