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Question

$\sin 2x$ + $4\tan x$ = $9$ $\frac{\sqrt{3}}{2}$

Toán Cấp 3 · Answer 1 · 5/27/2016 10:20:09 PM

Điều kiện:

$cos(x)\neq 0\Rightarrow x\neq \frac{\pi}{2}+k\pi(k\in Z)$

ptr ban đầu:

$2sin(x)cos(x)+\frac{4sin(x)}{cos(x)}=\frac{9\sqrt{3}}{2}$

$\Leftrightarrow 4sin(x)cos^2(x)+8sin(x)=9\sqrt{3}cos(x)$

$\Leftrightarrow 4tan(x)+8tan(x)[1+tan^2(x)]=9\sqrt{3}[1+tan^2(x)](1)$ do (

$cos(x)\neq 0$ )

Đặt

$t=tan(x)$ ,

$(1)$ thành:

$8t^3-9\sqrt{3}t^2+12t-9\sqrt{3}=0$

$\Leftrightarrow (t-\sqrt{3})(8t^2-\sqrt{3}t+9)=0$

$\Leftrightarrow t=\sqrt{3}\Rightarrow tan(x)=\sqrt{3}\Rightarrow x=\frac{\pi}{3}+k\pi(k\in Z)$

Vậy ptrlg có nghiệm là

$x=\frac{\pi}{3}+k\pi(k\in Z)$