Đáp án:
Giải thích các bước giải:
Do $\pi <x< \dfrac{3\pi}{2}$ nên $Sinx,cos x<0$
$Cosx=\sqrt{\dfrac{1}{tan^2x+1}}=-\dfrac{1}{\sqrt{8}}$
$Sinx =\sqrt{1-\dfrac{1}{8}}=-\dfrac{\sqrt{14}}{4}$
Ta có :
$D=sin(x-3\pi)\\D=sinx.cos3\pi -sin3\pi .cos x\\=\dfrac{\sqrt{14}}{4}$
$E=2cos^2x-1-sinx\\=2.\dfrac{-1}{8}-1-\dfrac{-\sqrt{14}}{4}\\=-\dfrac{5+\sqrt{14}}{4}$
$F=tan(\dfrac{\pi}{6}+x)\\=\dfrac{tan\dfrac{\pi}{6}+tanx}{1-tan\dfrac{\pi}{6}.tanx}\\=\dfrac{\dfrac{3\sqrt{7}+\sqrt{3}}{3}}{\dfrac{3-\sqrt{21}}{3}}\\=\dfrac{3\sqrt{7}+\sqrt{3}}{3-\sqrt{21}}$
