cot x – 1 = cos2x/(1+tanx) +sin x^{2} -1/2 . sin2x mn giải giúp nha cam on nhieu 28/09/2021 Bởi Genesis cot x – 1 = cos2x/(1+tanx) +sin x^{2} -1/2 . sin2x mn giải giúp nha cam on nhieu
Đáp án: $x = \dfrac{\pi }{4} + k\pi $ $(k\in\mathbb Z)$ Lời giải: $\begin{array}{l} \cot x – 1 = \dfrac{{\cos 2x}}{{1 + \tan x}} + {\sin ^2}x – \dfrac{1}{2}\sin 2x\\ \Leftrightarrow \dfrac{{\cos x}}{{\sin x}} – 1 = \dfrac{{\cos 2x}}{{1 + \dfrac{{\sin x}}{{\cos x}}}} + {\sin ^2}x – \dfrac{1}{2}\sin 2x\\ \Leftrightarrow \dfrac{{\cos x}}{{\sin x}} – 1 = \left( {\cos x – \sin x} \right)\cos x + {\sin ^2}x – \sin x\cos x\\ \Leftrightarrow \dfrac{{\cos x – \sin x}}{{\sin x}} = \left( {\cos x – \sin x} \right)\cos x + \sin \left( {\cos x – \sin x} \right)\\ \Leftrightarrow \left( {\cos x – \sin x} \right)\left[ {\dfrac{1}{{\sin x}} – \cos x + \sin x} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos x – \sin x = 0\\ 1 – \cos x\sin x + {\sin ^2}x = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \cos x = \sin x\\ 1 – \dfrac{1}{2}\sin 2x + \dfrac{{1 – \cos 2x}}{2} = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \tan x = 1\\ 2 – \sin 2x + 1 – \cos 2x = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \tan x = 1\\ \sin 2x + \cos 2x = 3\text{ (vô nghiệm)}\end{array} \right. \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \end{array}$ $(k\in\mathbb Z)$ Bình luận
Đáp án:
$x = \dfrac{\pi }{4} + k\pi $ $(k\in\mathbb Z)$
Lời giải:
$\begin{array}{l} \cot x – 1 = \dfrac{{\cos 2x}}{{1 + \tan x}} + {\sin ^2}x – \dfrac{1}{2}\sin 2x\\ \Leftrightarrow \dfrac{{\cos x}}{{\sin x}} – 1 = \dfrac{{\cos 2x}}{{1 + \dfrac{{\sin x}}{{\cos x}}}} + {\sin ^2}x – \dfrac{1}{2}\sin 2x\\ \Leftrightarrow \dfrac{{\cos x}}{{\sin x}} – 1 = \left( {\cos x – \sin x} \right)\cos x + {\sin ^2}x – \sin x\cos x\\ \Leftrightarrow \dfrac{{\cos x – \sin x}}{{\sin x}} = \left( {\cos x – \sin x} \right)\cos x + \sin \left( {\cos x – \sin x} \right)\\ \Leftrightarrow \left( {\cos x – \sin x} \right)\left[ {\dfrac{1}{{\sin x}} – \cos x + \sin x} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos x – \sin x = 0\\ 1 – \cos x\sin x + {\sin ^2}x = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \cos x = \sin x\\ 1 – \dfrac{1}{2}\sin 2x + \dfrac{{1 – \cos 2x}}{2} = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \tan x = 1\\ 2 – \sin 2x + 1 – \cos 2x = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \tan x = 1\\ \sin 2x + \cos 2x = 3\text{ (vô nghiệm)}\end{array} \right. \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \end{array}$ $(k\in\mathbb Z)$