Giải phương trình:    $  3\sqrt{\tan x+1}.(\sin x+2\cos x)=5(\sin x+\cos x)               (1)$
Điều kiện  $\begin{cases}\cos x \neq 0\\ \tan x\geq -1\end{cases}$
Đặt   $ t=\tan x \geq -1$
Khi đó PT $(1)$      $\Leftrightarrow    3\sqrt{\tan x+1}=5.\frac{\sin x+3\cos x}{\sin x+2\cos x}$ (  $\sin x+2\cos x\neq 0$ do $\tan x\neq -2$)
$\Leftrightarrow 3\sqrt{\tan x+1}=5.\frac{\tan x+3}{\tan x+2}$ ( do $\cos x\neq 0$)
$\Leftrightarrow 3.\sqrt{t+1}=5.\frac{t+3}{t+2}             (2)$
Đặt : $f(t)=3.\sqrt{t+1}- 5.\frac{t+3}{t+2}  $
       
Ta có :   $f'(t)=\frac{3}{2\sqrt{t+1}}+ \frac{5}{(t+2)^2} > 0 , \forall t > -1$
$\Rightarrow f(t)$ đồng biến trên $ (-1;+\infty)$, mà $ f(3)=0$

Do đó phương trình $(2)$ có nghiệm duy nhất    $t=3 \Leftrightarrow   \tan x=3=\tan\alpha  \Leftrightarrow  x=\alpha+ k\pi  , k \in Z$

Thẻ

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