a, ĐK: \(x\ne\dfrac{5\pi}{6}+k2\pi;x\ne\dfrac{\pi}{6}+k2\pi\)

\(\dfrac{2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)}{2sinx-1}=-1\)

\(\Leftrightarrow2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)=1-2sinx\)

\(\Leftrightarrow-cos\left(3x-\dfrac{\pi}{2}\right)+\sqrt{3}cos^3x.\dfrac{cos^2x-3sin^2x}{cos^2x}=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(cos^2x-3sin^2x\right)=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(4cos^2x-3\right)=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cos3x=-2sinx\)

\(\Leftrightarrow\dfrac{1}{2}sin3x-\dfrac{\sqrt{3}}{2}cos3x-sinx=0\)

\(\Leftrightarrow sin\left(3x-\dfrac{\pi}{3}\right)-sinx=0\)

\(\Leftrightarrow2cos\left(2x-\dfrac{\pi}{6}\right)sin\left(x-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(2x-\dfrac{\pi}{6}\right)=0\\sin\left(x-\dfrac{\pi}{6}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\\x-\dfrac{\pi}{6}=k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)

Đối chiếu điều kiện ta được:

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\\x=\dfrac{7\pi}{6}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(3tan\cdot\left(4x-\dfrac{\pi}{3}\right)+\sqrt{3}=0\)

\(\Rightarrow tan\left(4x-\dfrac{\pi}{3}\right)=-\dfrac{1}{\sqrt{3}}=tan\left(-\dfrac{\pi}{6}\right)\)

\(\Rightarrow4x-\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k\pi\)\(\Rightarrow4x=\dfrac{\pi}{6}+k\pi\)

\(\Rightarrow x=\dfrac{\pi}{24}+k\dfrac{\pi}{4}\left(k\in Z\right)\)

ĐK: \(x\ne k\pi\)

\(3tan^2\left(x-\dfrac{\pi}{2}\right)=2.\dfrac{1-sinx}{sinx}\)

\(\Leftrightarrow3cot^2x=\dfrac{2}{sinx}-2\)

\(\Leftrightarrow\dfrac{3}{sin^2x}-3=\dfrac{2}{sinx}-2\)

\(\Leftrightarrow\dfrac{3}{sin^2x}-\dfrac{2}{sinx}-1=0\)

\(\Leftrightarrow\left(\dfrac{1}{sinx}-1\right)\left(\dfrac{3}{sinx}+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{sinx}-1=0\\\dfrac{3}{sinx}+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-3\left(l\right)\end{matrix}\right.\)

\(sinx=1\Leftrightarrow x=\dfrac{\pi}{2}+\text{k}2\pi\left(tm\right)\)

Vậy phương trihf đã cho có nghiệm \(x=\dfrac{\pi}{2}+\text{k}2\pi\)

ĐKXĐ: \(x\ne k\pi\)

\(3cot^2x=2\left(\dfrac{1-sinx}{sinx}\right)\)

\(\Leftrightarrow\dfrac{3cos^2x}{sin^2x}=2\left(\dfrac{1-sinx}{sinx}\right)\)

\(\Leftrightarrow\dfrac{3\left(1-sinx\right)\left(1+sinx\right)}{sin^2x}-2\left(\dfrac{1-sinx}{sinx}\right)=0\)

\(\Leftrightarrow\left(\dfrac{1-sinx}{sinx}\right)\left(\dfrac{3+3sinx}{sinx}-2\right)=0\)

\(\Leftrightarrow\left(\dfrac{1-sinx}{sinx}\right)\left(\dfrac{3+sinx}{sinx}\right)=0\)

\(\Leftrightarrow sinx=1\)

\(\Rightarrow x=\dfrac{\pi}{2}+k2\pi\)

Lời giải:

$\tan (\frac{\pi}{2}+x)-3\tan ^2x=\frac{\cos 2x-1}{\cos ^2x}=\frac{2\cos ^2x-2}{\cos ^2x}=\frac{2(\cos ^2x-1)}{\cos ^2x}$

$=\frac{-2\sin ^2x}{\cos ^2x}=-2\tan ^2x$

$\Leftrightarrow \tan (x+\frac{\pi}{2})=\tan ^2x$

Dễ thấy $\tan x=0$ không thỏa mãn nên $\tan x\neq 0$. Do đó pt $\Leftrightarrow \tan ^2x=\tan [\pi +(x-\frac{\pi}{2})]=\tan (x-\frac{\pi}{2})=-\tan (\frac{\pi}{2}-x)=-\cot x =\frac{-1}{\tan x}$

$\Rightarrow \tan ^3x=-1$

$\Rightarrow \tan x=-1$

$\Rightarrow x=\frac{-\pi}{4}+k\pi$ với $k$ nguyên.

a/

\(\Leftrightarrow tan2x=-tan40^0\)

\(\Leftrightarrow tan2x=tan\left(-40^0\right)\)

\(\Rightarrow2x=-40^0+k180^0\)

\(\Rightarrow x=-20^0+k90^0\)

b/

\(\Leftrightarrow tan\left(2x-15^0\right)=1\)

\(\Rightarrow2x-15^0=45^0+k180^0\)

\(\Rightarrow x=30^0+k90^0\)

c/

\(\Leftrightarrow tan\left(60^0-x\right)=-\frac{1}{\sqrt{3}}\)

\(\Rightarrow60^0-x=-30^0+k180^0\)

\(\Rightarrow x=90^0+k180^0\)

d/

\(\Leftrightarrow tan\left(3x+\frac{2\pi}{5}\right)=-tan\left(\frac{\pi}{5}\right)\)

\(\Leftrightarrow tan\left(3x+\frac{2\pi}{5}\right)=tan\left(-\frac{\pi}{5}\right)\)

\(\Rightarrow3x+\frac{2\pi}{5}=-\frac{\pi}{5}+k\pi\)

\(\Rightarrow x=-\frac{\pi}{5}+\frac{k\pi}{3}\)

a, y xác định `<=> 3cos(2x+3) \ne 0`

`<=>cos(2x+3) \ne 0`

`<=>2x+3 \ne π/2+kπ`

`<=>x \ne π/4 -3/2 +k π/2 (k \in ZZ)`

b, y xác định `<=> sin(x/3+π/4) \ne0`

`<=> x/3+π/4 \ne kπ`

`<=> x \ne (-3π)/4+ k3π`

ĐKXĐ: 

a.

\(cos\left(2x+3\right)\ne0\)

\(\Leftrightarrow2x+3\ne\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=-\dfrac{3}{2}+\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

b.

\(sin\left(\dfrac{x}{3}+\dfrac{\pi}{4}\right)\ne0\)

\(\Leftrightarrow\dfrac{x}{3}+\dfrac{\pi}{4}\ne k\pi\)

\(\Leftrightarrow x\ne-\dfrac{3\pi}{4}+k3\pi\)

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