b: \(\sin^2x=cos^22x+cos^23x\)
=>\(\frac{1-cos2x}{2}=\frac{1+cos4x}{2}+\frac{1+cos6x}{2}\)
=>1-cos2x=1+cos4x+1+cos6x
=>cos6x+cos4x+cos2x+1=0
=>\(\left(cos6x+cos2x\right)+\left(cos4x+1\right)=0\)
=>\(2\cdot cos4x\cdot cos2x+2\cdot cos^22x=0\)
=>\(cos2x\left(cos4x+cos2x\right)=0\)
TH1: cos2x=0
=>\(2x=\frac{\pi}{2}+k\pi\)
=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)
TH2: \(cos4x+cos2x=0\)
=>\(2\cdot cos3x\cdot cosx=0\)
=>\(cos3x\cdot cosx=0\)
=>\(\left[\begin{array}{l}cos3x=0\\ cosx=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=\frac{\pi}{2}+k\pi\\ x=\frac{\pi}{2}+k\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{6}+\frac{k\pi}{3}\\ x=\frac{\pi}{2}+k\pi\end{array}\right.\)
d: \(\sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\frac14\)
=>\(\sin^4x+\sin^4\left(\frac{\pi}{4}-x\right)=\frac14\)
=>\(\frac{3-4\cdot cos2x+cos4x}{8}+\frac{3-4\cdot cos\left(\frac{\pi}{2}-2x\right)+cos\left(\pi-4x\right)}{8}=\frac14\)
=>\(\frac{3-4\cdot cos2x+cos4x}{8}+\frac{3-4\cdot\sin2x-cos4x}{8}=\frac14\)
=>\(6-4\cdot cos2x-4\cdot\sin2x=\frac14\cdot8=2\)
=>3-2cos2x-2sin2x=1
=>cos2x+sin2x=1
=>\(\sqrt2\cdot\sin\left(2x+\frac{\pi}{4}\right)=1\)
=>\(\sin\left(2x+\frac{\pi}{4}\right)=\frac{1}{\sqrt2}\)
=>\(\left[\begin{array}{l}2x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\ 2x+\frac{\pi}{4}=\pi-\frac{\pi}{4}+k2\pi=\frac34\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=k2\pi\\ 2x=\frac12\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=k\pi\\ x=\frac14\pi+k\pi\end{array}\right.\)
