a: ĐKXĐ: 2*sin x+1<>0

=>sin x<>-1/2

=>x<>-pi/6+k2pi và x<>7/6pi+k2pi

b: ĐKXĐ: \(\dfrac{1+cosx}{2-cosx}>=0\)

mà 1+cosx>=0

nên 2-cosx>=0

=>cosx<=2(luôn đúng)

c ĐKXĐ: tan x>0

=>kpi<x<pi/2+kpi

d: ĐKXĐ: \(2\cdot cos\left(x-\dfrac{pi}{4}\right)-1< >0\)

=>cos(x-pi/4)<>1/2

=>x-pi/4<>pi/3+k2pi và x-pi/4<>-pi/3+k2pi

=>x<>7/12pi+k2pi và x<>-pi/12+k2pi

e: ĐKXĐ: x-pi/3<>pi/2+kpi và x+pi/4<>kpi

=>x<>5/6pi+kpi và x<>kpi-pi/4

f: ĐKXĐ: cos^2x-sin^2x<>0

=>cos2x<>0

=>2x<>pi/2+kpi

=>x<>pi/4+kpi/2

1.

ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\tanx-sinx\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\\dfrac{sinx}{cosx}-sinx\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\sinx\ne0\\cosx\ne1\end{matrix}\right.\) \(\Leftrightarrow sin2x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{2}\)

2.

ĐKXĐ: \(sin2x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{2}\)

3. 

ĐKXĐ: \(\left\{{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)\ne0\\cos\left(x-\dfrac{\pi}{4}\right)\ne0\end{matrix}\right.\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{2}\right)\ne0\Leftrightarrow cos2x\ne0\)

\(\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

câu 2 ..... \(\dfrac{cos^22x}{sin^22x}=cot^22x\) nên suy ra sin2x khác 0 đúng hơm

còn câu 3, tui ko hiểu chỗ sin(2x-pi/4).. sao ở đây rớt xuống dợ

a.

\(\left\{{}\begin{matrix}sin\left(3x+\dfrac{\pi}{6}\right)\ne0\\cos2x\ne0\\sinx\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{\pi}{18}+\dfrac{k\pi}{3}\\x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x\ne-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

b.

Do \(5+2cot^2x-sinx=4+2cot^2x+\left(1-sinx\right)>0\) nên hàm xác định khi:

\(\left\{{}\begin{matrix}sinx\ne0\\sin\left(x+\dfrac{\pi}{2}\right)\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\end{matrix}\right.\) \(\Leftrightarrow sin2x\ne0\)

\(\Leftrightarrow x\ne\dfrac{k\pi}{2}\)

1. \(sin\left(\dfrac{\pi}{3}-x\right)\ne0\Leftrightarrow\dfrac{\pi}{3}-x\ne k\pi\Leftrightarrow x\ne\dfrac{\pi}{3}-k\pi\)

2. \(cos2x\ne0\Leftrightarrow2x\ne\dfrac{\pi}{2}+k\pi\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

3. \(\sqrt{1+sinx}-\sqrt{2}\ge0\Leftrightarrow1+sinx\ge2\Leftrightarrow sinx\ge1\Leftrightarrow sinx=1\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

4. \(\sqrt{2-2cosx}-2\ne0\Leftrightarrow2-2cosx\ne4\Leftrightarrow cosx\ne-1\Leftrightarrow x\ne\pi+k2\pi\)

5. \(1-\sqrt{1+sin3x}\ne0\Leftrightarrow sin3x\ne0\Leftrightarrow3x\ne k\pi\Leftrightarrow x\ne\dfrac{k\pi}{3}\)

\(\sqrt{3}sinx+cosx\ne0\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\ne0\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)\ne0\)

\(\Leftrightarrow x+\dfrac{\pi}{6}\ne k\pi\)

\(\Leftrightarrow x\ne-\dfrac{\pi}{6}+k\pi\)

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.

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

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

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

1.

Hàm số xác định khi \(\left\{{}\begin{matrix}\dfrac{1+x}{1-x}\ge0\\1-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1\le x< 1\\x\ne1\end{matrix}\right.\Leftrightarrow-1\le x< 1\)

2.

Hàm số xác định khi \(cosx+1\ne0\Leftrightarrow cosx\ne-1\Leftrightarrow x\ne-\pi+k2\pi\)

3.

Hàm số xác định khi \(cosx-cos3x\ne0\Leftrightarrow sin2x.sinx\ne0\Leftrightarrow\left[{}\begin{matrix}x\ne k\pi\\x\ne\dfrac{k\pi}{2}\end{matrix}\right.\)

1.

\(2sin\left(x+10^o\right)-\sqrt{12}cos\left(x+10^o\right)=3\)

\(\Leftrightarrow\dfrac{1}{2}sin\left(x+10^o\right)-\dfrac{\sqrt{3}}{2}cos\left(x+10^o\right)=\dfrac{3}{4}\)

\(\Leftrightarrow sin\left(x+50^o\right)=\dfrac{3}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+50^o=arcsin\left(\dfrac{3}{4}\right)+k360^o\\x+50^o=180^o-arcsin\left(\dfrac{3}{4}\right)+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-50^o+arcsin\left(\dfrac{3}{4}\right)+k360^o\\x=130^o-arcsin\left(\dfrac{3}{4}\right)+k360^o\end{matrix}\right.\)

2.

\(\sqrt{3}sin4x-cos4x=\sqrt{3}\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin4x-\dfrac{1}{2}cos4x=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow sin\left(4x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\4x-\dfrac{\pi}{3}=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\pi}{12}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

3.

\(sin2x-cot\dfrac{\pi}{5}.cos2x=1\)

\(\Leftrightarrow\sqrt{1+cot\dfrac{\pi}{5}}\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}sin2x-\dfrac{cot\dfrac{\pi}{5}}{\sqrt{1+cot\dfrac{\pi}{5}}}.cos2x\right)=1\)

\(\Leftrightarrow sin\left[2x-arccos\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)\right]=\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-arccos\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)=arcsin\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)+k2\pi\\2x-arccos\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)=\pi-arcsin\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}arccos\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)+\dfrac{1}{2}arcsin\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)+k\pi\\x=\dfrac{\pi}{2}+\dfrac{1}{2}arccos\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)-\dfrac{1}{2}arcsin\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)+k\pi\end{matrix}\right.\)

ĐK: \(\left\{{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)\ne0\\sin^4x-cos^4x\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{\pi}{4}\ne k\pi\\\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+k\pi\\cos2x\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+k\pi\\2x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+k\pi\\x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)