a/ \(\tan^2x-\cot^2\left(x-\frac{\pi}{4}\right)=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-1-\frac{1}{\sin^2\left(x-\frac{\pi}{4}\right)}+1=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\sin x.\cos\frac{\pi}{4}-\cos x.\sin\frac{\pi}{4}\right)^2}=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\frac{\sqrt{2}}{2}\sin x-\frac{\sqrt{2}}{2}\cos x\right)^2}=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x}=0\)

\(\Leftrightarrow\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x-\cos^2x=0\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}\cos^2x-\sin x.\cos x-\frac{1}{2}\cos^2x=0\)

\(\Leftrightarrow\cos^2x+\sin x.\cos x-\frac{1}{2}=0\)

Đến đây là dễ r nha bn :3

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.

\(\left(2sinx-1\right)\left(2cos2x+2sinx+1\right)=3-4\left(1-sin^2x\right)\)

\(\Leftrightarrow\left(2sinx-1\right)\left(2cos2x+2sinx+1\right)=4sin^2x-1\)

\(\Leftrightarrow\left(2sinx-1\right)\left(2cos2x+2sinx+1\right)-\left(2sinx-1\right)\left(2sinx+1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(2cos2x+2sinx+1-2sinx-1\right)=0\)

\(\Leftrightarrow2cos2x\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sinx=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

2.

ĐKXĐ: ...

\(\Leftrightarrow cot\left(\frac{\pi}{4}-x\right)=-\frac{1}{\sqrt{3}}\)

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

\(\Leftrightarrow x=\frac{7\pi}{12}+k\pi\)

3.

\(\Leftrightarrow cos\frac{x}{4}sinx+sin\frac{x}{4}.cosx-3\left(sin^2x+cos^2x\right)+cosx=0\)

\(\Leftrightarrow sin\left(x+\frac{x}{4}\right)=-cosx\)

\(\Leftrightarrow sin\frac{5x}{4}=sin\left(x-\frac{\pi}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{5x}{4}=x-\frac{\pi}{2}+k2\pi\\\frac{5x}{4}=\frac{3\pi}{2}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

4.

\(\Leftrightarrow2sinx.cosx-\left(1-2sin^2x\right)+3sinx-cosx-1=0\)

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

\(\Leftrightarrow cosx\left(2sinx-1\right)+\left(2sinx-1\right)\left(sinx+2\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(sinx+cosx+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2sinx-1=0\\sinx+cosx=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

a/ \(cosx>0\Rightarrow cosx=\sqrt{1-sin^2x}=\frac{4}{5}\)

\(\Rightarrow tanx=-\frac{3}{4}\Rightarrow A=\frac{129}{20}\)

b/ \(B=\frac{5sinx+3cosx}{3cosx-2sinx}=\frac{\frac{5sinx}{sinx}+\frac{3cosx}{sinx}}{\frac{3cosx}{sinx}-\frac{2sinx}{sinx}}=\frac{5+3cotx}{3cotx-2}=\frac{5+9}{9-2}\)

c/ \(C=\frac{sinx.cosx\left(cotx-2tanx\right)}{sinx.cosx\left(5cotx+tanx\right)}=\frac{cos^2x-2sin^2x}{5cos^2x+sin^2x}=\frac{cos^2x-2\left(1-cos^2x\right)}{5cos^2x+1-cos^2x}=\frac{3cos^2x-2}{4cos^2x+1}=...\)

d/ Không dịch được đề, ko biết mẫu số bên trái nó đến đâu cả

a/ ĐKXĐ: \(sinx\ne-1\)

\(\Leftrightarrow\left(2sinx+1\right)\left(3cos4x+2sinx\right)+4cos^2x+1=8+8sinx\)

\(\Leftrightarrow6sinx.cos4x+4sin^2x+3cos4x+2sinx+4cos^2x+1=8+8sinx\)

\(\Leftrightarrow6sinx.cos4x+3cos4x-6sinx-3=0\)

\(\Leftrightarrow6sinx\left(cos4x-1\right)+3\left(cos4x-1\right)=0\)

\(\Leftrightarrow\left(6sinx+3\right)\left(cos4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\cos4x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\1-2sin^22x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sin^2x\left(1-sin^2x\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sin^2x\left(1+sinx\right)\left(1-sinx\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=0\\sinx=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=k\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

b/ ĐKXĐ: \(\left\{{}\begin{matrix}tanx\ne-1\\cosx\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left(1+sinx+cos2x\right).\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=cosx\left(1+\frac{sinx}{cosx}\right)\)

\(\Leftrightarrow\left(1+sinx+cos2x\right)\left(sinx+cosx\right)=cosx+sinx\)

\(\Leftrightarrow\left(cosx+sinx\right)\left(sinx+cos2x\right)=0\)

\(\Leftrightarrow sinx+cos2x=0\)

\(\Leftrightarrow-2sin^2x+sinx+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\left(l\right)\\sinx=-\frac{1}{2}\end{matrix}\right.\)

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

a/

\(\Leftrightarrow4sinx.cosx\left(sin^4x-cos^4x\right)=sin^24x\)

\(\Leftrightarrow2sin2x\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin^24x\)

\(\Leftrightarrow-2sin2x.cos2x=sin^24x\)

\(\Leftrightarrow-sin4x=sin^24x\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin4x=-1\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{4}\\x=-\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)

b/

\(\Leftrightarrow2\left(1-cosx\right)-\sqrt{3}cos2x=1+1+cos\left(2x-\frac{3\pi}{2}\right)\)

\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=sin\left(2\pi-2x\right)\)

\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=-sin2x\)

\(\Leftrightarrow sin2x-\sqrt{3}cos2x=2cosx\)

\(\Leftrightarrow\frac{1}{2}sin2x-\sqrt{3}cos2x=cosx\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=cosx=sin\left(\frac{\pi}{2}-x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{5}{4}=0\)

\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{3}\right)-\frac{5}{4}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\\sin\left(x+\frac{\pi}{3}\right)=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(y=4\left(1-sin^2x\right)+2sinx+2=-4sin^2x+2sinx+6\)

Đặt \(sinx=t\in\left[-1;1\right]\Rightarrow y=f\left(t\right)=-4t^2+2t+6\)

\(-\dfrac{b}{2a}=\dfrac{1}{4}\in\left[-1;1\right]\)

\(f\left(-1\right)=0\) ; \(f\left(\dfrac{1}{4}\right)=\dfrac{25}{4}\); \(f\left(1\right)=4\)

\(\Rightarrow y_{max}=\dfrac{25}{4}\) khi \(sinx=\dfrac{1}{4}\)

\(y_{min}=0\) khi \(sinx=-1\)

Ta có: \(y=4cos^2x+2sinx+2=4-4sin^2x+2sinx+2=-4sin^2x+2sinx+6=-\left(4sin^2x-2sinx+\dfrac{1}{16}-\dfrac{1}{16}-6\right)=-\left(2sin^2x-\dfrac{1}{4}\right)^2+\dfrac{97}{16}\)

Ta có: \(-\left(2sin^2x-\dfrac{1}{4}\right)^2\le0\Rightarrow y\le\dfrac{97}{16}\)

Vậy \(y_{max}=\dfrac{97}{16}\)