pt luong giac lop 11

J

jet_nguyen

Gợi ý:
ĐK:....
$$ \tan x+\cot x=4$$$$\Longleftrightarrow \tan^2x-4\tan x+1=0$$$$ \Longleftrightarrow \left[\begin{array}{1} \tan x= 2+\sqrt{3} \\ \tan x= 2-\sqrt{3} \end{array}\right.$$$$ \Longleftrightarrow \left[\begin{array}{1} x= \dfrac{5 \pi}{12}+k\pi \\ x= \dfrac{ \pi}{12}+k\pi \end{array}\right.$$ $\bullet$ $x= \dfrac{5 \pi}{12}+k\pi $ $(k \in Z)$mà $ x \in (0,2\pi)$ nên:
$$0< \dfrac{5 \pi}{12}+k\pi < 2\pi$$$$\Longleftrightarrow \dfrac{-5}{12}<k<\dfrac{1}{12}$$$$\Longleftrightarrow k=0,1$$ Vậy: $ x= \dfrac{5 \pi}{12}, \dfrac{17 \pi}{12}$
$\bullet$ Tương tự với trường hợp 2.
 
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H

hocviencsnd

[TEX]tanx+cotx=4[/TEX]
[TEX]\Leftrightarrow \frac{2}{sin2x}=4\Leftrightarrow sin2x=\frac{1}{2}[/TEX]
[TEX]\Leftrightarrow \left\{\begin{matrix}x=\frac{\prod }{12}+k\prod & \\x=\frac{5\prod }{12}+k\prod & \end{matrix}\right.[/TEX]

Đến đây bạn tự xét nghiệm với ĐK nha ...:D
 
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