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: \(sin4x\ne0\Leftrightarrow x\ne\frac{k\pi}{4}\)
\(\Leftrightarrow\frac{sin4x}{cosx}+\frac{sin4x}{sin2x}=2\)
\(\Leftrightarrow4sinx.cos2x+2cos2x=2\)
\(\Leftrightarrow cos2x\left(2sinx+1\right)=1\)
\(\Leftrightarrow\left(1-2sin^2x\right)\left(2sinx+1\right)=1\)
\(\Leftrightarrow4sin^3x+2sin^2x-2sinx=0\)
\(\Leftrightarrow2sinx\left(2sin^2x+sinx-1\right)=0\)
\(\Leftrightarrow2sin^2x+sinx-1=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
