\(4\left(sin^4x+cos^4x\right)+sin4x-2=0\)

\(\Leftrightarrow4\left(1-2sin^2x.cos^2x\right)+2sin2x.cos2x-2=0\)

\(\Leftrightarrow2-2sin^22x+2sin2x.cos2x=0\)

\(\Leftrightarrow2\left(1-sin^22x+sin2x.cos2x\right)=0\)

\(\Leftrightarrow2\left(cos^22x+sin2x.cos2x\right)=0\)

\(\Leftrightarrow2cos2x\left(cos2x+sin2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x+sin2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2};x=-\dfrac{\pi}{8}+\dfrac{k\pi}{2}\)

\(cos^3xsinx-sin^3xcosx=sinx.cosx\left(cos^2x-sin^2x\right)=\dfrac{1}{2}sin2x.cos2x=\dfrac{1}{4}sin4x\)

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

\(=1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{3}{4}+\dfrac{1}{4}cos4x=\dfrac{1}{4}\left(3+cos4x\right)\)

\(\text{a) }cos^2x+sin2x-1=0\\ \Leftrightarrow2sinx\cdot cosx-sin^2x=0\\ \Leftrightarrow sinx\left(2cosx-sinx\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=2cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=0\\tanx=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=a\pi\\x=arctan\left(2\right)+b\pi\end{matrix}\right.\)

\(\text{b) }\sqrt{3}sin2x+cos^4x-sin^4x=\sqrt{2}\\ \Leftrightarrow\sqrt{3}sin2x+\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=\sqrt{2}\\ \Leftrightarrow\frac{\sqrt{3}}{2}\cdot sin2x+\frac{1}{2}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sin\frac{\pi}{4}\\ \\ \Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=\frac{\pi}{4}+a2\pi\\2x+\frac{\pi}{6}=\frac{3\pi}{4}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{24}+a\pi\\x=\frac{7\pi}{24}+b\pi\end{matrix}\right.\)

\(c\text{) }cos^2x-sin^2x=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\\ \Leftrightarrow cos^2x-sin^2x=\sqrt{2}\left(sinx\cdot\frac{\sqrt{2}}{2}+cosx\cdot\frac{\sqrt{2}}{2}\right)\\ \Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)=sinx+cosx\\ \Leftrightarrow\left[{}\begin{matrix}cosx-sinx=1\\sinx=-cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cos^2x+\left(cosx-1\right)^2=1\\tanx=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\\tanx=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+a\pi\\x=b2\pi\\x=\frac{3\pi}{4}=c\pi\end{matrix}\right.\)

\(d\text{) }4\left(sin^4x+cos^4x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow4\left(1-2sin^2x\cdot cos^2x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow-8sin^2x\cdot cos^2x+\sqrt{3}sin4x=-2\\ \Leftrightarrow-2sin^22x+\sqrt{3}sin4x=-2\\ \Leftrightarrow cos4x-1+\sqrt{3}sin4x=-2\\ \Leftrightarrow\frac{1}{2}cos4x+\frac{\sqrt{3}}{2}sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\frac{\pi}{6}\cdot cos4x+cos\frac{\pi}{6}\cdot sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\left(4x+\frac{\pi}{6}\right)=sin\frac{-\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{6}=\frac{-\pi}{6}+a2\pi\\4x+\frac{\pi}{6}=\frac{7\pi}{6}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-\pi}{12}+\frac{a\pi}{2}\\x=\frac{\pi}{4}+\frac{b\pi}{2}\end{matrix}\right.\)

\(e\text{) }4sinx\cdot cosx\cdot cos2x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x\cdot\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}cos4x=1\\ \Leftrightarrow sin4x\cdot cos\frac{\pi}{4}+cos4x\cdot sin\frac{\pi}{4}=1\\ \Leftrightarrow sin\left(4x+\frac{\pi}{4}\right)=1=sin\frac{\pi}{2}\\ \Leftrightarrow4x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{\pi}{16}+\frac{k\pi}{2}\)

\(\Leftrightarrow4\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x+\sqrt{3}sin4x=2\)

\(\Leftrightarrow4-2sin^22x+\sqrt{3}sin4x=2\)

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

\(\Leftrightarrow\frac{1}{2}cos4x+\frac{\sqrt{3}}{2}sin4x=-\frac{1}{2}\)

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

\(\Leftrightarrow...\)

\(\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=sin4x\)

\(\Leftrightarrow cos^2x-sin^2x=sin4x\)

\(\Leftrightarrow cos2x=sin4x=cos\left(\frac{\pi}{2}-4x\right)\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{3}\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)

\(2cos^2x-1=sin6x\)

\(\Leftrightarrow cos2x=sin6x=cos\left(\frac{\pi}{2}-6x\right)\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{16}+\frac{k\pi}{4}\\x=\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)

\(2\left(cos^2x-1\right)=sinx.cos3x\)

\(\Leftrightarrow-2sin^2x=sinx.cos3x\)

\(\Leftrightarrow sinx.cos3x+2sin^2x=0\)

\(\Leftrightarrow sinx\left(cos3x+2sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos3x+2sinx=0\left(1\right)\end{matrix}\right.\)

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 \(a\in\left[0;\pi\right]\)

\(\Rightarrow sin2x.cosa+cos2x.sina=0\)

\(\Leftrightarrow sin\left(2x+a\right)=0\)

\(\Rightarrow2x+a=k\pi\Rightarrow x=-\frac{a}{2}+\frac{k\pi}{2}\)