Toán Lớp 11: Cos7x-cos3x=2sin2x Giải hộ với ạ cần gấp

Question

Toán Lớp 11:  Cos7x-cos3x=2sin2x Giải hộ với ạ cần gấp

tuminh25 · Answer

$ begin{array}{l}  cos 7x -  cos 3x = 2 sin 2x     Leftrightarrow  - 2 sin 5x sin 2x = 2 sin 2x     Leftrightarrow 2 sin 2x + 2 sin 5x sin 2x = 0     Leftrightarrow 2 sin 2x left( { sin 5x + 1}  right) = 0     Leftrightarrow  left[  begin{array}{l}  sin 2x = 0    sin 5x =  - 1  end{array}  right.  Leftrightarrow  left[  begin{array}{l} 2x = k pi    5x =  -  dfrac{ pi }{2} + k2 pi   end{array}  right.     Leftrightarrow  left[  begin{array}{l} x =  dfrac{{k pi }}{2}   x =  -  dfrac{ pi }{{10}} +  dfrac{{k2 pi }}{5}  end{array}  right. left( {k  in Z}  right)  end{array}$

minhtuquynh · Answer

~rai~        ( cos7x- cos3x=2 sin2x   Leftrightarrow -2 sin5x sin2x=2 sin2x   Leftrightarrow 2 sin2x+2 sin5x sin2x=0   Leftrightarrow 2 sin2x(1+ sin5x)=0   Leftrightarrow  left[ begin{array}{I} sin2x=0   sin5x=-1 end{array} right.   Leftrightarrow  left[ begin{array}{I}2x=k pi  5x=- dfrac{ pi}{2}+k2 pi end{array} right.   Leftrightarrow  left[ begin{array}{I}x=k dfrac{ pi}{2}  x=- dfrac{ pi}{10}+k dfrac{2 pi}{5}. end{array} right. quad(k in mathbb{Z})   text{Vậy S=} left {k dfrac{ pi}{2};- dfrac{ pi}{10}+k dfrac{2 pi}{5} Big|k in mathbb{Z} right }. )