Đáp án:

a) $x=\{k\pi;k\dfrac{\pi}3\}$

b) $x=\{\dfrac{\pi}4+k\pi;-\dfrac{\pi}2+k\pi\}$

c) $x=\{\dfrac{\pi}6+k2\pi;\dfrac{5\pi}6+k2\pi\}$

$(k\in\mathbb Z)$

Lời giải:

a)

$\tan x+\tan2x=\tan3x$

$\Rightarrow\dfrac{\sin x}{\cos x}+\dfrac{\sin 2x}{\cos 2x}=\dfrac{\sin 3x}{\cos 3x}$

Đk

$\left\{ \begin{array}{l} \cos x\ne0 \\ \cos 2x\ne0 \\\cos 3x\ne0\end{array} \right .\Leftrightarrow \left\{ \begin{array}{l} x\ne\dfrac{\pi}{2}+k\pi\\ 2x\ne\dfrac{\pi}{2}+k\pi\\3x\ne\dfrac{\pi}{2}+k\pi\end{array} \right .$

$\Leftrightarrow \left\{ \begin{array}{l} x\ne\dfrac{\pi}{2}+k\pi\\ x\ne\dfrac{\pi}{4}+k\dfrac{\pi}{2}\\x\ne\dfrac{\pi}{6}+k\dfrac{\pi}{3}\end{array} \right .$

Phương trình tương đương:

$\Rightarrow\dfrac{\sin x\cos 2x+\cos x\sin2x}{\cos x\cos 2x}=\dfrac{\sin 3x}{\cos 3x}$

$\Rightarrow \dfrac{\sin3x}{\cos x\cos 2x}=\dfrac{\sin 3x}{\cos 3x}$

$\Rightarrow \sin 3x(\dfrac{1}{\cos x\cos 2x}-\dfrac{1}{\cos 3x})=0$

TH1: $\Rightarrow\sin 3x=0\Rightarrow 3x= k\pi$

$\Rightarrow x= k\dfrac{\pi}{3}(k\in\mathbb Z)$ (nhận)

TH2: $\dfrac{1}{\cos x\cos 2x}-\dfrac{1}{\cos 3x}=0$

$\Rightarrow \cos 3x-\cos x\cos 2x=0$ $\Rightarrow \cos 3x-\dfrac{1}{2}(\cos x+\cos 3x)=0$

$\Rightarrow \dfrac{1}{2}\cos 3x-\dfrac{1}{2}\cos x=0$

$\Rightarrow \cos 3x-\cos x=0$

$\Rightarrow \cos 3x=\cos x$

$\Rightarrow 3x=\pm x+k2\pi$

$\Rightarrow x=k\pi$ (nhận) hoặc $x=\dfrac{k\pi}2$ (loại) $(k\in\mathbb Z)$

Vậy phương trình có nghiệm $x=\{k\pi;k\dfrac{\pi}3\}$ $(k\in\mathbb Z)$

b) $\sin x+\cos x=\dfrac1{\sin x}$

Đk:

$\sin x\ne 0\Leftrightarrow x\ne k\pi(k\in\mathbb Z)$

Phương trình tương đương

$\sqrt 2\sin (x+\dfrac{\pi}{4})\sin x=1$

$\Rightarrow\sqrt2\dfrac{1}{2}[\cos\dfrac{\pi}{4}-\cos(2x+\dfrac{\pi}{4})]=1$

$\Rightarrow \dfrac{1}{2}-\dfrac{1}{\sqrt2}\cos(2x+\dfrac{\pi}{4})=1$

$\Rightarrow \cos(2x+\dfrac{\pi}{4})=\dfrac{-1}{\sqrt2}$

$\Rightarrow 2x+\dfrac{\pi}{4}=\pm\dfrac{3\pi}{4}+k2\pi$

$\Rightarrow x=\dfrac{\pi}4+k\pi$ hoặc $x=-\dfrac{\pi}2+k\pi$ $(k\in\mathbb Z)$

c)

$(2\sin x-\cos x)(1+\cos x)=\sin^2x$

$\Leftrightarrow(2\sin x-\cos x)(1+\cos x)=1-\cos^2x=(1-\cos x)(1+\cos x)$

Trường hợp 1: $1+\cos x=0$

$\Leftrightarrow\cos x=-1\Leftrightarrow x=\pi+k2\pi$ $(k\in\mathbb Z)$

Trường hợp 2: $2\sin x-\cos x-1+\cos x=0$

$\Leftrightarrow 2\sin x=1\Leftrightarrow \sin x=\dfrac12$

$\Leftrightarrow x=\dfrac{\pi}6+k2\pi$ hoặc $x=\dfrac{5\pi}6+k2\pi$ $(k\in\mathbb Z)$.