a/

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

\(sin^4x-cos^4x=\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=\left(sinx-cosx\right)\left(sinx+cosx\right)\)

Do đó pt tương đương:

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

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

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

Xét (1), đặt \(sinx+cosx=a\Rightarrow sinx.cosx=\frac{a^2-1}{2}\) với \(\left|a\right|\le\sqrt{2}\)

\(\left(1\right)\Leftrightarrow2+2a+\frac{a^2-1}{2}=0\)

\(\Leftrightarrow a^2+4a+3=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=-3\left(l\right)\end{matrix}\right.\)

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

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

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

b/

Nhận thấy \(sinx=0\) không phải nghiệm, pt tương đương:

\(sinx.cosx.cos2x.cos4x.cos8x=\frac{1}{16}sinx\)

\(\Leftrightarrow8sin2x.cos2x.cos4x.cos8x=sinx\)

\(\Leftrightarrow4sin4x.cos4x.cos8x=sinx\)

\(\Leftrightarrow2sin8x.cos8x=sinx\)

\(\Leftrightarrow sin16x=sinx\)

\(\Rightarrow\left[{}\begin{matrix}16x=x+k2\pi\\16x=\pi-x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{15}\\x=\frac{\pi}{17}+\frac{k2\pi}{17}\end{matrix}\right.\)

c/

ĐKXĐ: \(sin4x\ne0\Leftrightarrow x\ne\frac{k\pi}{4}\)

\(\Leftrightarrow\frac{sin4x}{cosx}+\frac{sin4x}{sin2x}=2\)

\(\Leftrightarrow4sinx.cos2x+2cos2x=2\)

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

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

\(\Leftrightarrow4sin^3x+2sin^2x-2sinx=0\)

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

\(\Leftrightarrow2sin^2x+sinx-1=0\)

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

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

\(\Leftrightarrow\left(1+cos^2x\right)\left(sinx+cosx\right)=sin^2x+cos^2x+2sinx.cosx\)

\(\Leftrightarrow\left(1+cos^2x\right)\left(sinx+cosx\right)=\left(sinx+cosx\right)^2\)

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

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

Xét (1), đặt \(\left\{{}\begin{matrix}sinx=a\\cosx=b\end{matrix}\right.\) với \(\left|a\right|;\left|b\right|\le1\) ta được hệ:

\(\left\{{}\begin{matrix}a^2+b^2=1\\1+b^2-a-b=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\a=b^2-b+1\end{matrix}\right.\)

\(\Rightarrow b^2+\left(b^2-b+1\right)^2=1\)

\(\Leftrightarrow b\left(b^3-2b^2+4b-2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=0\Rightarrow a=1\Rightarrow x=\frac{\pi}{2}+k2\pi\\b^3-2b^2+4b+2=0\left(2\right)\end{matrix}\right.\)

Pt (2) là 1 pt ko giải được theo kiến thức phổ thông

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

\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)

b, \(VT=\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}=\frac{cos2x}{\left(sin^2x+cos^2x\right)^2-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{1-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{cos^2x-2.sin^2x.cos^2x}\)

=\(\frac{cos2x}{cos^2x.\left(1-2.sin^2x\right)}=\frac{cos2x}{cos^2x.cos2x}=\frac{1}{cos^2x}=1+tan^2x=VP\)

d, \(VT=\left(\frac{cosx}{1+sinx}+tanx\right).\left(\frac{sinx}{1+cosx}+cotx\right)=\left(\frac{cosx}{1+sinx}+\frac{sinx}{cosx}\right).\left(\frac{sinx}{1+cosx}+\frac{cosx}{sinx}\right)\)

\(=\left(\frac{cos^2x+sinx.\left(1+sinx\right)}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx.\left(1+cosx\right)}{sinx.\left(1+cosx\right)}\right)=\left(\frac{cos^2x+sinx+sin^2x}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx+cos^2x}{sinx.\left(1+cosx\right)}\right)\)

=\(\frac{1}{cosx.sinx}=VP\)

e, \(VT=cos^2x.\left(cos^2x+2sin^2x+sin^2x.tan^2x\right)=cos^2x.\left(1+sin^2x.\left(1+tan^2x\right)\right)=cos^2x.\left(1+tan^2x\right)=cos^2x.\frac{1}{cos^2x}=1=VP\)

c, \(VT=\frac{sin^2x}{cosx.\left(1+tanx\right)}-\frac{cos^2x}{sinx.\left(1+cosx\right)}=\frac{sin^3x.\left(1+cosx\right)-cos^3x.\left(1+tanx\right)}{sinx.cosx.\left(1+tanx\right).\left(1+cosx\right)}\)

=\(\frac{sin^3x+sin^3x.cotx-cos^3x-cos^3.tanx}{\left(sinx+cosx\right)^2}=\frac{sin^3x+sin^2xcosx-cos^3x-cos^2sinx}{\left(sinx+cosx\right)^2}=\frac{sin^2x.\left(sinx+cosx\right)-cos^2x.\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}\)

\(=\frac{\left(sin^2x-cos^2x\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=\frac{\left(sinx-cosx\right).\left(sinx+cosx\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=sinx-cosx=VP\)

Đây nha bạn

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

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

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

\(=cos^4\frac{x}{2}+cos^2\frac{x}{2}\)

a.

\(\Leftrightarrow\left(1+cos4x\right)sin2x=\frac{1}{2}\left(1+cos4x\right)\)

\(\Leftrightarrow\left(1+cos4x\right)\left(sin2x-\frac{1}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-1\\sin2x=\frac{1}{2}\end{matrix}\right.\)

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

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

b.

\(\Leftrightarrow cosx+sin^2x.cosx+sinx+cos^2x.sinx=sin^2x+cos^2x+2sinx.cosx\)

\(\Leftrightarrow sinx+cosx+sinx.cosx\left(sinx+cosx\right)=\left(sinx+cosx\right)^2\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=1\\sinx=1\end{matrix}\right.\)

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