Đáp án:

$\left\{\begin{array}{I}x=-\dfrac{\pi}4+k\pi\\x=\arctan\dfrac{3+\sqrt{17}}4+k\pi\end{array}\right.$ $(k\in\mathbb Z)$

Giải thích các bước giải:

$1+3\sin2x=2\tan x$

Điều kiện $\cos x\ne0\Leftrightarrow x\ne\dfrac{\pi}2+k\pi$ $(k\in\mathbb Z)$

Phương trình tương đương:

$1+6\sin x\cos x=2\tan x$

$\Leftrightarrow\dfrac1{\cos^2x}+6\tan x=2\tan x.dfrac1{\cos^2x}$

$\Leftrightarrow(1+\tan^2x)+6\tan x=2\tan x(1+\tan^2x)$

$\Leftrightarrow2\tan^3x-\tan^2x-4\tan x-1=0$

$\Leftrightarrow(\tan x+1)(2\tan^2x-3\tan x-1)=0$

$\Leftrightarrow\left[\begin{array}{I}\tan x=-1\\\tan x=\dfrac{3\pm\sqrt{17}}4\end{array}\right.$

$\Leftrightarrow\left[\begin{array}{I}x=-\dfrac{\pi}4+k\pi\\x=\arctan\dfrac{3+\sqrt{17}}4+k\pi\end{array}\right.$ $(k\in\mathbb Z)$

Vậy $\left\{\begin{array}{I}x=-\dfrac{\pi}4+k\pi\\x=\arctan\dfrac{3+\sqrt{17}}4+k\pi\end{array}\right.$ $(k\in\mathbb Z)$.

Đáp án:

$x\in\{\dfrac34\pi+k\pi,\dfrac{8k\pi-\pi +4\arcsin \left(\dfrac{\sqrt{2}}{6}\right)}{8}, \dfrac{3\pi +8k\pi-4\arcsin \left(\dfrac{\sqrt{2}}{6}\right)}{8}\}$

Giải thích các bước giải:

ĐKXĐ: $x\ne \dfrac{\pi}{2}+k\pi, k\in Z$

Ta có:

$1+3\sin2x=2\tan x$

$\to 3+3\sin2x=2\tan x+2$

$\to 3\left(1+\sin2x\right)=2\left(\tan x+1\right)$

$\to 3\left(\sin^2x+\cos^2x+2\sin x\cos x\right)=2\left(\dfrac{\sin x}{\cos x}+1\right)$

$\to 3\left(\sin x+\cos x\right)^2=2\cdot\dfrac{\sin x+\cos x}{\cos x}$

$\to 3\left(\sin x+\cos x\right)^2-2\cdot\dfrac{\sin x+\cos x}{\cos x}=0$

$\to \left(\sin x+\cos x\right)\left(3\left(\sin x+\cos x\right)-\dfrac{2}{\cos x}\right)=0$

$\to \left(\sin x+\cos x\right)\left(3\left(\sin x\cos x+\cos^2x\right)-2\right)=0$

$\to \left(\sin x+\cos x\right)\left(\dfrac32\left(2\sin x\cos x+2\cos^2x-1+1\right)-2\right)=0$

$\to \left(\sin x+\cos x\right)\left(\dfrac32\left(\sin2x+\cos2x+1\right)-2\right)=0$

$\to \left(\sin x+\cos x\right)\left(3\left(\sin2x+\cos2x+1\right)-4\right)=0$

$\to \left(\sin x+\cos x\right)\left(3\left(\sin2x+\cos2x\right)-1\right)=0$

$\to \sin x+\cos x=0$

$\to \sin x=-\cos x$

$\to \dfrac{\sin x}{\cos x}=-1$

$\to \tan x=-1$

$\to x=\dfrac34\pi+k\pi$

Hoặc $3\left(\sin2x+\cos2x\right)-1=0$

$\to \sin2x+\cos2x=\dfrac13$

$\to \sin2x\cdot\dfrac{\sqrt{2}}{2}+\cos2x\cdot\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{6}$

$\to \sin2x\cdot\cos\left(\dfrac{\pi}{4}\right)+\cos2x\cdot\sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{6}$

$\to \sin\left(2x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{6}$

$\to 2x+\dfrac{\pi}{4}=\arcsin\dfrac{\sqrt{2}}{6}+k2\pi\to x=\dfrac{8k\pi-\pi +4\arcsin \left(\dfrac{\sqrt{2}}{6}\right)}{8}$

Hoặc $2x+\dfrac{\pi}{4}=\pi-\arcsin\dfrac{\sqrt{2}}{6}+k2\pi\to x=\dfrac{3\pi +8k\pi-4\arcsin \left(\dfrac{\sqrt{2}}{6}\right)}{8}$