a, Ta có : \(\sin^2x+\cos^2x=1\)
\(\Rightarrow\sin x=\sqrt{1-\cos^2x}=\left|\dfrac{\sqrt{15}}{4}\right|\)
Mà \(0< x< \dfrac{\pi}{2}\)
\(\Rightarrow\sin x=\dfrac{\sqrt{15}}{4}\)
Ta lại có : \(\left\{{}\begin{matrix}\sin2x=2\sin x\cos x=\dfrac{\sqrt{15}}{8}\\\cos2x=2\cos^2x-1=-\dfrac{7}{8}\end{matrix}\right.\)
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c, Ta có : \(\tan2x=\dfrac{2\tan x}{1-\tan^2x}=\dfrac{4}{3}=\dfrac{\sin2x}{\cos2x}\)
- Ta có HPT : \(\left\{{}\begin{matrix}\sin^22x+\cos^22x=1\\3\sin2x-4\cos2x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sin2x=\left|\dfrac{4}{5}\right|\\\cos2x=\left|\dfrac{3}{5}\right|\end{matrix}\right.\)
Lại có : \(\pi< x< \dfrac{3}{2}\pi\)
\(\Rightarrow\left\{{}\begin{matrix}\sin2x=\dfrac{4}{5}\\\cos2x=\dfrac{3}{5}\end{matrix}\right.\)
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