cotx. cot2x= -1 Cot( 5pi/3 -3pi) -tan(2x+pi/3)=0

Đáp án + Giải thích các bước giải:

 $\cot x . \cot 2x = -1$

$\Leftrightarrow \dfrac{\cos x}{\sin x} . \dfrac{\cos 2x}{\sin 2x} = -1$
$\Leftrightarrow \dfrac{\cos x}{\sin x} . \dfrac{1 - 2\sin^2 x}{2 \sin x\cos x} = -1$

$\Leftrightarrow \dfrac{1}{\sin x} . \dfrac{1 - 2\sin^2 x}{2\sin x} = -1$

$\Leftrightarrow \dfrac{1 - 2\sin^2 x}{2\sin^2 x} = -1$

$\Leftrightarrow \dfrac{1 - 2\sin^2 x + 2\sin^2 x}{2\sin^2 x} = 0$

$\Leftrightarrow \dfrac{1}{2\sin^2 x} = 0$(vô lý)

Vậy phương trình vô nghiệm

$\cot \bigg(\dfrac{5\pi}{3} - 3\pi\bigg) - \tan \bigg(2x + \dfrac{\pi}{3}\bigg) = 0$

$\Leftrightarrow \tan \bigg(\dfrac{\pi}{2} - \dfrac{5\pi}{3} + 3\pi\bigg) = \tan\bigg(2x + \dfrac{\pi}{3}\bigg)$

$\Leftrightarrow \tan \bigg(\dfrac{11\pi}{6}\bigg) = \tan\bigg(2x + \dfrac{\pi}{3}\bigg)$
$\Leftrightarrow \tan \bigg(\dfrac{-\pi}{6}\bigg) = \tan\bigg(2x + \dfrac{\pi}{3}\bigg)$

$\Leftrightarrow 2x + \dfrac{\pi}{3} = \dfrac{-\pi}{6} + k\pi$

$\Leftrightarrow 2x = \dfrac{-\pi}{2} + k\pi$

$\Leftrightarrow x = \dfrac{-\pi}{4} + k\dfrac{\pi}{2}$