\(\frac{sin^22x+4sin^2x-4}{1-8sin^2x-cos4x}=\frac{4sin^2x.cos^2x-4\left(1-sin^2x\right)}{1-8sin^2x-\left(1-2sin^22x\right)}=\frac{4sin^2x.cos^2x-4cos^2x}{2sin^22x-8sin^2x}\)

\(=\frac{-4cos^2x\left(1-sin^2x\right)}{8sin^2x.cos^2x-8sin^2x}=\frac{-4cos^2x.cos^2x}{-8sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{2sin^4x}=\frac{1}{2}cot^4x\)

\(\frac{cos2x}{cot^2x-tan^2x}=\frac{cos2x.sin^2x.cos^2x}{cos^4x-sin^4x}=\frac{\left(cos^2x-sin^2x\right).\left(2sinx.cosx\right)^2}{4\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)}=\frac{1}{4}sin^22x\)

1a.

Đặt \(5x+6=u\)

\(cos2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+6=\dfrac{\pi}{4}+k2\pi\\5x+6=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

1b.

Đặt \(2x+1=u\)

\(cos2u+3sinu=2\)

\(\Leftrightarrow1-2sin^2u+3sinu=2\)

\(\Leftrightarrow2sin^2u-3sinu+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=1\\sinu=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x+1\right)=1\\sin\left(2x+1\right)=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{\pi}{2}+k2\pi\\2x+1=\dfrac{\pi}{6}+k2\pi\\2x+1=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)

2a.

\(cos^2x-sin^2x+sin^2x+2cosx+1=0\)

\(\Leftrightarrow cos^2x+2cosx+1=0\)

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

\(\Leftrightarrow cosx=-1\)

\(\Leftrightarrow x=\pi+k2\pi\)

\(\frac{sin^22x-4sin^2x}{sin^22x-4\left(1-sin^2x\right)}=\frac{4sin^2x.cos^2x-4sin^2x}{4sin^2x.cos^2x-4cos^2x}=\frac{sin^2x\left(cos^2x-1\right)}{cos^2x\left(sin^2x-1\right)}=\frac{-sin^4x}{-cos^4x}=tan^4x\)

ĐKXĐ: \(x\ne\frac{k\pi}{2}\)

\(\frac{4sin^2x.cos^2x-4sin^2x}{4sin^2x.cos^2x+4sin^2x}+1=2tan^2x\)

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

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

Đặt \(cos^2x=t\Rightarrow0< t< 1\)

\(\Rightarrow\frac{t}{t+1}=\frac{1-t}{t}\Leftrightarrow t^2=1-t^2\Leftrightarrow t^2=\frac{1}{2}\)

\(\Leftrightarrow t=\frac{\sqrt{2}}{2}\Leftrightarrow cos^2x=\frac{\sqrt{2}}{2}\)

ĐKXĐ: \(cosx\ne0\)

\(\frac{sin^22x-2}{4sin^2x.cos^2x-4cos^2x}=\frac{sin^2x}{cos^2x}\)

\(\Leftrightarrow\frac{sin^22x-2}{4cos^2x\left(sin^2x-1\right)}=\frac{sin^2x}{cos^2x}\)

\(\Leftrightarrow\frac{2-sin^22x}{4cos^4x}=\frac{sin^2x}{cos^2x}\)

\(\Leftrightarrow2-sin^22x=4sin^2x.cos^2x\)

\(\Leftrightarrow2-sin^22x=sin^22x\)

\(\Leftrightarrow sin^22x=1\Leftrightarrow cos2x=0\)

\(\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)

a. ĐKXĐ: ...

\(\frac{sinx}{cosx}+\frac{sin2x}{cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow\frac{sin2x.cosx+cos2x.sinx}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow\frac{sin3x}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow sin3x\left(\frac{cosx.cos2x+cos3x}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{cosx\left(2cos^2x-1\right)+4cos^3x-3cosx}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{6cos^2x-4}{cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{3cos2x-1}{cos2x.cos3x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cos2x=\frac{1}{3}\end{matrix}\right.\)

b.

\(cos2x\left(2cos^22x-1\right)=\frac{1}{2}\)

\(\Leftrightarrow4cos^32x-2cos2x-1=0\)

Pt bậc 3 này ko giải được, chắc bạn ghi nhầm đề

c. ĐKXĐ: ...

\(\frac{cosx}{sinx}-\frac{sinx}{cosx}=cosx-sinx\)

\(\Leftrightarrow\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx.cosx}=cosx-sinx\)

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

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

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)

\(\Rightarrow t=\frac{t^2-1}{2}\Rightarrow t^2-2t-1=0\Rightarrow\left[{}\begin{matrix}t=1+\sqrt{2}\left(l\right)\\t=1-\sqrt{2}\end{matrix}\right.\)

\(\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1-\sqrt{2}\Rightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}}\Rightarrow x=...\)

d.

\(\Leftrightarrow2\left(cosx-cos3x\right)sin4x=sin3x\)

\(\Leftrightarrow2sin4x.cosx-2sin4x.cos3x=sin3x\)

\(\Leftrightarrow sin5x+sin3x-sin7x-sinx=sin3x\)

\(\Leftrightarrow sin5x-sin7x-sinx=0\)

\(\Leftrightarrow-2cos6x.sinx-sinx=0\)

\(\Leftrightarrow sinx\left(2cos6x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos6x=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

a/

\(\Leftrightarrow4sin^3x+6\sqrt{2}sinx.cosx-8sinx=0\)

\(\Leftrightarrow2sinx\left(2sin^2x+3\sqrt{2}cosx-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\2sin^2x+3\sqrt{2}cosx-4=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2\left(1-cos^2x\right)+3\sqrt{2}cosx-4=0\)

\(\Leftrightarrow-2cos^2x+3\sqrt{2}cosx-2=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=\sqrt{2}>1\left(l\right)\\cosx=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow x=\pm\frac{\pi}{4}+k2\pi\)

b/

\(\Leftrightarrow4cos^3x+8sinx.cosx-7cosx=0\)

\(\Leftrightarrow cosx\left(4cos^2x+8sinx-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\4cos^2x+8sinx-7=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow4\left(1-sin^2x\right)+8sinx-7=0\)

\(\Leftrightarrow-4sin^2x+8sinx-3=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=\frac{3}{2}\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)

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

c/

ĐKXĐ; ...

\(\Leftrightarrow\frac{sinx}{cosx}+\frac{cosx}{sinx}-5+\frac{3}{sin^22x}=0\)

\(\Leftrightarrow\frac{sin^2x+cos^2x}{sinx.cosx}-5+\frac{3}{sin^22x}=0\)

\(\Leftrightarrow\frac{3}{sin^22x}+\frac{2}{sin2x}-5=0\)

Đặt \(\frac{1}{sin2x}=t\Rightarrow3t^2+2t-5=0\)

\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{5}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{1}{sin2x}=1\\\frac{1}{sin2x}=-\frac{5}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{3}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{1}{2}arcsin\left(-\frac{3}{5}\right)+k\pi\\x=\frac{\pi}{2}-\frac{1}{2}arcsin\left(-\frac{3}{5}\right)+k\pi\end{matrix}\right.\)

a

\(\Leftrightarrow\left(3sinx-sin3x\right)cos3x+\left(3cosx+cos3x\right)sin3x+3\sqrt{3}cos4x=3\)

\(\Leftrightarrow\left(sinx.cos3x+sin3x.cosx\right)+\sqrt{3}cos4x=1\)

\(\Leftrightarrow sin4x+\sqrt{3}cos4x=1\)

Tới đây thôi, mình lười ghi rồi =))

b

\(\Leftrightarrow\left(1-cos2x\right)\left(2sin^2x-1\right)\left(2sin^2+1\right)=cos2x\left(7cos^22x+3cos2x-4\right)\)

\(\Leftrightarrow\left(1-cos2x\right)\left(-cos2x\right)\left(2-cos2x\right)=cos2x\left(7cos^22x+3cos2x+4\right)\)

\(\Leftrightarrow-cos^22x+3cos2x-2=7cos^22x+3cos2x+4\)

\(\Leftrightarrow4cos^22x+3=0\)

=> pt vô nghiệm