giải pt 11/ $sin3x+cos3x-sin+cosx= √2cos2x$

<=>√2sin(3x+$\frac{π}{4}$) -√2sin(x-$\frac{π}{4}$) =√2cos2x

<=>2cos2xsin(x+$\frac{π}{4}$ )=cos2x

TH1 cos2x=0

<=>0=0(luôn đúng)

<=>x=$\frac{π}{4}$ +$\frac{kπ}{2}$ (k∈Z)

TH2 cos2x$\neq$ 0

<=>2sin(x+$\frac{π}{4}$ )=1

<=>\(\left[ \begin{array}{l}x=$\frac{-π}{12}$ +k2π\\x=$\frac{7π}{12}$+k2π \end{array} \right.\) (k∈Z)