GIẢI PHƯƠNG TRÌNH BẬC CAO SỬ DỤNG CÔNG THỨC TAN a. 4sin2x + tanx= 3√3 b. 2cos2x – tanx = 1 c. tanx.tan2x= 3√3 d. sin2x + 2cotx= 3

Đáp án:

$a) \, x = \dfrac{\pi}{3} + k\pi \quad (k\in \Bbb Z)$

$b) \, \left[\begin{array}{l}x = \dfrac{3\pi}{4} + k\pi\\x = \arctan(-1 + \sqrt2) + k\pi\\x = \arctan(-1 – \sqrt2) + k\pi\end{array}\right.\quad (k \in \Bbb Z)$

$c) \, \left[\begin{array}{l}x = \arctan\left(\sqrt{\dfrac{3\sqrt3}{2 + 3\sqrt3}}\right) + k\pi\\x = \arctan\left(-\sqrt{\dfrac{3\sqrt3}{2 + 3\sqrt3}}\right) + k\pi\end{array}\right.\quad (k \in \Bbb Z)$

$d) \, x = \dfrac{3\pi}{4} + k\pi \quad (k \in \Bbb )$

Giải thích các bước giải:

$\begin{array}{l}a) \, 4\sin2x + \tan x = 3\sqrt3\qquad (a)\\ ĐKXĐ: \, \cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi}{2} + n\pi\\ (a) \Leftrightarrow 8\sin x\cos x + \tan x = 3\sqrt3\\ \Leftrightarrow 8\tan x + \dfrac{\tan x}{\cos^2x} = \dfrac{3\sqrt3}{\cos^2x}\\ \Leftrightarrow 8\tan x + \tan x(1 + \tan^2x) = 3\sqrt3(1 + \tan^2x)\\ \Leftrightarrow \tan^3x – 3\sqrt3\tan^2x + 9\tan x – 3\sqrt3 = 0\\ \Leftrightarrow (\tan x – \sqrt3)^3 = 0\\ \Leftrightarrow \tan = \sqrt3\\ \Leftrightarrow x = \dfrac{\pi}{3} + k\pi \quad (k\in \Bbb Z)\\ b)\, 2\cos2x – \tan x = 1\qquad (b)\\ ĐKXĐ: \, \cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi}{2} + n\pi\\ (b) \Leftrightarrow 2(1 – 2\sin^2x) – \tan x = 1\\ \Leftrightarrow 4\sin^2x + \tan x – 1 = 0\\ \Leftrightarrow 4\tan^2x + \dfrac{\tan x}{\cos^2x} – \dfrac{1}{\cos^2x}=0\\ \Leftrightarrow 4\tan^2x + \tan x(1 + \tan^2x) – (1 + \tan^2x) = 0\\ \Leftrightarrow \tan^3x + 3\tan^2x + \tan x – 1 = 0\\ \Leftrightarrow \left[\begin{array}{l}\tan x = – 1\\\tan x = -1 + \sqrt2\\\tan x = -1 – \sqrt2\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{3\pi}{4} + k\pi\\x = \arctan(-1 + \sqrt2) + k\pi\\x = \arctan(-1 – \sqrt2) + k\pi\end{array}\right.\quad (k \in \Bbb Z)\\ c)\, \tan x.\tan2x = 3\sqrt3\qquad (c)\\ ĐKXĐ: \, \begin{cases}\cos x \ne 0\\\cos2x \ne 0\end{cases} \Leftrightarrow \begin{cases}x \ne \dfrac{\pi}{2} +n\pi\\x \ne \dfrac{\pi}{4} +n\dfrac{\pi}{2}\end{cases}\quad (n \in \Bbb Z)\\ (c)\Leftrightarrow \tan x.\dfrac{2\tan x}{1 – \tan^2x} = 3\sqrt3\\ \Leftrightarrow 2\tan^2x = 3\sqrt3(1 – \tan^2x)\\ \Leftrightarrow (2 + 3\sqrt3)\tan^2x = 3\sqrt3\\ \Leftrightarrow \tan^2x = \dfrac{3\sqrt3}{2 + 3\sqrt3}\\ \Leftrightarrow \tan x = \pm \sqrt{\dfrac{3\sqrt3}{2 + 3\sqrt3}}\\ \Leftrightarrow \left[\begin{array}{l}x = \arctan\left(\sqrt{\dfrac{3\sqrt3}{2 + 3\sqrt3}}\right) + k\pi\\x = \arctan\left(-\sqrt{\dfrac{3\sqrt3}{2 + 3\sqrt3}}\right) + k\pi\end{array}\right.\quad (k \in \Bbb Z)\\ d)\, \sin2x + 2\cot x + 3\qquad (d)\\ ĐKXĐ:\, \sin x \ne 0 \Leftrightarrow x \ne n\pi \quad (n \in \Bbb Z)\\ \text{Nhận thấy $\cos x = 0$ không là nghiệm của phương trình}\\ \text{Chia 2 vế của phương trình cho $\cos^2x$ ta được:}\\ (d) \Leftrightarrow 2\tan x + \dfrac{2}{\tan x.\cos^2x} + \dfrac{3}{\cos^2x} = 0\\ \Leftrightarrow 2\tan x + \dfrac{2}{\tan x}(1 + \tan^2x) + 3(1 + \tan^2x) = 0\\ \Leftrightarrow 2\tan^2x + 2(1 + \tan^2x) + 3\tan x(1 + \tan^2x) = 0\\ \Leftrightarrow 3\tan^3x + 4\tan^2x + 3\tan x + 2 = 0\\ \Leftrightarrow (\tan x + 1)(3\tan^2x + x + 2) = 0\\ \Leftrightarrow \tan x = -1\\ \Leftrightarrow x = \dfrac{3\pi}{4} + k\pi \quad (k \in \Bbb ) \end{array}$