Mn có ai biết lầm câu 7 ko chỉ tớ với

Đáp án + Giải thích các bước giải:

Bài 7:

$\cos 2x = -\dfrac{1}{3}$
$\Rightarrow \sin^2 2x = 1 - \cos^2 2x = 1 - \dfrac{1}{9} = \dfrac{8}{9}$

Ta có: $\pi < x < \dfrac{3\pi}{2} \Rightarrow 2\pi < 2x < 3\pi$

$\Rightarrow \sin 2x > 0$
$\Leftrightarrow \sin 2x = \dfrac{2\sqrt{2}}{3}$

Mặt khác, $\cos 2x = -\dfrac{1}{3} \Rightarrow 2 \cos^2 x - 1 = -\dfrac{1}{3}$

$\Leftrightarrow 2 \cos^2 x = \dfrac{2}{3}$
$\Leftrightarrow \cos^2 x = \dfrac{1}{3}$

$\Leftrightarrow \cos x = \dfrac{-\sqrt{3}}{3}\bigg(\pi < x < \dfrac{\pi}{2}\bigg)$

$\cos 4x = 2 \cos^2 2x - 1$

$= 2\bigg(-\dfrac{1}{3}\bigg)^2 - 1 = \dfrac{-7}{9}$

$\tan \bigg(2x - \dfrac{\pi}{4}\bigg) = \dfrac{\tan 2x - \tan \frac{\pi}{4}}{1 + \tan 2x \tan \frac{\pi}{4}}$

$= \dfrac{\tan 2x - 1}{1 + \tan 2x}$

$= \dfrac{\sin 2x - \cos 2x}{\sin 2x + \cos 2x}$

$= \dfrac{\frac{2\sqrt{2}}{3} + \frac{1}{3}}{\frac{2\sqrt{2}}{3} - \frac{1}{3}}$

$= \dfrac{2\sqrt{2} + 1}{2\sqrt{2} - 1}$