\(\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=sin4x\)
\(\Leftrightarrow cos^2x-sin^2x=sin4x\)
\(\Leftrightarrow cos2x=sin4x=cos\left(\frac{\pi}{2}-4x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-4x+k2\pi\\2x=4x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{3}\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
\(2cos^2x-1=sin6x\)
\(\Leftrightarrow cos2x=sin6x=cos\left(\frac{\pi}{2}-6x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-6x+k2\pi\\2x=6x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{16}+\frac{k\pi}{4}\\x=\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)
\(2\left(cos^2x-1\right)=sinx.cos3x\)
\(\Leftrightarrow-2sin^2x=sinx.cos3x\)
\(\Leftrightarrow sinx.cos3x+2sin^2x=0\)
\(\Leftrightarrow sinx\left(cos3x+2sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos3x+2sinx=0\left(1\right)\end{matrix}\right.\)
Bạn có ghi nhầm đề ko nhỉ, pt (1) dù giải được nhưng khá khó đấy, phải vận dụng công thức nhân 3 và nghiệm ko hề đẹp
\(cos^4x+sin^4x=1+\frac{1}{2}sin4x\)
\(\Leftrightarrow\left(cos^2x+sin^2x\right)^2-2\left(sinx.cosx\right)^2=1+\frac{1}{2}sin4x\)
\(\Leftrightarrow1-\frac{1}{2}sin^22x=1+\frac{1}{2}sin4x\)
\(\Leftrightarrow sin4x+sin^22x=0\)
\(\Leftrightarrow2sin2x.cos2x+sin^22x=0\)
\(\Leftrightarrow sin2x\left(2cos2x+sin2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=0\Rightarrow x=\frac{k\pi}{2}\\2cos2x+sin2x=0\left(1\right)\end{matrix}\right.\)
Xét (1)
\(\Leftrightarrow\frac{1}{\sqrt{5}}sin2x+\frac{2}{\sqrt{5}}cos2x=0\)
Đặt \(cosa=\frac{1}{\sqrt{5}}\) với \(a\in\left[0;\pi\right]\)
\(\Rightarrow sin2x.cosa+cos2x.sina=0\)
\(\Leftrightarrow sin\left(2x+a\right)=0\)
\(\Rightarrow2x+a=k\pi\Rightarrow x=-\frac{a}{2}+\frac{k\pi}{2}\)
