Toán giải pt 11/ $sin3x+cos3x-sin+cosx= √2cos2x$ 10/07/2021 By Clara 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) Trả lời
<=>√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)