ĐKXĐ: \(cosx\ne0\)

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

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

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

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

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

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

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

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

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

a.

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

\(\Leftrightarrow2-\left(2sin\dfrac{x}{2}cos\dfrac{x}{2}\right)^2=1\)

\(\Leftrightarrow1-sin^2x=0\)

\(\Leftrightarrow cos^2x=0\)

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

b.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)

\(\Leftrightarrow16-12.sin^22x=7\)

\(\Leftrightarrow3-4sin^22x=0\)

\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)

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

\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)

c.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos^22x+\dfrac{1}{4}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=cos^22x+\dfrac{1}{4}\)

\(\Leftrightarrow3-3sin^22x=4cos^22x\)

\(\Leftrightarrow3=3\left(sin^22x+cos^22x\right)+cos^22x\)

\(\Leftrightarrow3=3+cos^22x\)

\(\Leftrightarrow cos2x=0\)

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

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

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

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

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

\(sin^2x+sin^22x=1\)

\(\Leftrightarrow2sin^2x-1+2sin^22x-2=-1\)

\(\Leftrightarrow-cos2x-2cos^22x+1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)

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

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

ĐKXĐ: \(x\ne\frac{k\pi}{2}\)

\(\Leftrightarrow\left(tanx+cotx\right)^2=\frac{4+sin4x}{sin^22x}+2\)

\(\Leftrightarrow\left(\frac{sin^2x+cos^2x}{\frac{1}{2}sin2x}\right)^2=\frac{4+sin4x+2sin^22x}{sin^22x}\)

\(\Leftrightarrow\frac{4}{sin^22x}=\frac{4+sin4x+2sin^22x}{sin^22x}\)

\(\Leftrightarrow2sin^22x+sin4x=0\)

\(\Leftrightarrow1-cos4x+sin4x=0\)

\(\Leftrightarrow\sqrt{2}cos\left(4x+\frac{\pi}{4}\right)=1\)

\(\Leftrightarrow cos\left(4x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)

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

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

\(\Leftrightarrow\frac{\cos^2x-4\sin^2x.\cos^2x}{4\cos^2x}=\frac{1}{2}\left(\cos\frac{\pi}{3}-\cos2x\right)\)

\(\Leftrightarrow1-4\sin^2x=2\left(\frac{1}{2}-\cos2x\right)\)

\(\Leftrightarrow1-4\sin^2x=1-2\cos2x\)

\(\Leftrightarrow2\sin^2x=\cos2x\)

\(\Leftrightarrow1-\cos2x=\cos2x\)

\(\Leftrightarrow\cos2x=\frac{1}{2}\Leftrightarrow x=\pm\frac{\pi}{6}+k\pi,k\in Z\) thỏa mãn điều kiện

a)\(pt\Leftrightarrow\frac{1-cos8x}{2}+\frac{1-cos6x}{2}=\frac{1-cos4x}{2}+\frac{1-cos2x}{2}\)

\(\Leftrightarrow cos2x+cos4x=cos6x+cos8x\)

\(\Leftrightarrow2cos3x\cdot cosx=2cos7x\cdot cosx\)

\(\Leftrightarrow2cos\left(cos3x-cos7x\right)=0\)

\(\Leftrightarrow2cosx\cdot\left(-2\right)\cdot sin5x\cdot sin\left(-2x\right)=0\)

\(\Leftrightarrow cosx\cdot sin2x\cdot sin5x=0\)

\(\Leftrightarrow sin2x\cdot sin5x=0\)(do sin2x=0 <=>2sinx*cosx=0 gồm th cosx=0 r`)

\(\Leftrightarrow\left[\begin{array}{nghiempt}sin2x=0\\sin5x=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{k\pi}{2}\\x=\frac{k\pi}{5}\end{array}\right.\)\(\left(k\in Z\right)\)

b)\(pt\Leftrightarrow1-cos2x+1-cos4x=1+cos6x+1+cos8x\)

\(\Leftrightarrow cos2x+cos8x+cos4x+cos6x=0\)

\(\Leftrightarrow cos10x\cdot cos6x+cos10x\cdot cos2x=0\)

\(\Leftrightarrow cos10x\left(cos6x+cos2x\right)=0\)

\(\Leftrightarrow cos10x\cdot cos8x\cdot cos4x=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}cos10x=0\\cos8x=0\\cos4x=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{\pi}{20}+\frac{k\pi}{10}\\x=\frac{\pi}{16}+\frac{k\pi}{8}\\x=\frac{\pi}{8}+\frac{k\pi}{4}\end{array}\right.\)