Toán Lớp 11: Giải phương trình sin(x-pi/4)=-cos(2x+pi/3)

Question

Toán Lớp 11:  Giải phương trình sin(x-pi/4)=-cos(2x+pi/3)

tuyethutanhthu · Answer

~rai~        ( sin left(x- dfrac{ pi}{4} right)=- cos left(2x+ dfrac{ pi}{3} right)   Leftrightarrow  cos left[ dfrac{ pi}{2}- left(x- dfrac{ pi}{4} right) right]= cos left[ pi- left(2x- dfrac{ pi}{3} right) right]   Leftrightarrow  cos left( dfrac{3 pi}{4}-x right)= cos left( dfrac{2 pi}{3}-2x right)   Leftrightarrow  left[ begin{array}{I} dfrac{3 pi}{4}-x= dfrac{2 pi}{3}-2x+k2 pi   dfrac{3 pi}{4}-x=2x- dfrac{2 pi}{3}+k2 pi end{array} right.   Leftrightarrow  left[ begin{array}{I}x=- dfrac{ pi}{12}+k2 pi  3x= dfrac{17 pi}{12}+k2 pi end{array} right.   Leftrightarrow  left[ begin{array}{I}x=- dfrac{ pi}{12}+k2 pi  x= dfrac{17 pi}{36}+k dfrac{2 pi}{3}. end{array} right. quad(k in mathbb{Z})   text{Vậy S=} left {- dfrac{ pi}{12}+k2 pi; dfrac{17 pi}{36}+k dfrac{2 pi}{3} Big|k in mathbb{Z} right }. )

landinhngoc97 · Answer

$ begin{array}{l}  sin  left( {x -  dfrac{ pi }{4}}  right) =  -  cos  left( {2x +  dfrac{ pi }{3}}  right)     Leftrightarrow  sin  left( {x -  dfrac{ pi }{4}}  right) =  cos  left( {2x +  pi  +  dfrac{ pi }{3}}  right)     Leftrightarrow  sin  left( {x -  dfrac{ pi }{4}}  right) =  cos  left( {2x +  dfrac{{4 pi }}{3}}  right)     Leftrightarrow  sin  left( {x -  dfrac{ pi }{4}}  right) =  cos  left[ { dfrac{ pi }{2} -  left( { -  dfrac{{5 pi }}{6} - 2x}  right)}  right]     Leftrightarrow  sin  left( {x -  dfrac{ pi }{4}}  right) =  sin  left( { -  dfrac{{5 pi }}{6} - 2x}  right)     Leftrightarrow  sin  left( {x -  dfrac{ pi }{4}}  right) =  sin  left( { -  dfrac{{5 pi }}{6} - 2x}  right)     Leftrightarrow  left[  begin{array}{l} x -  dfrac{ pi }{4} =  -  dfrac{{5 pi }}{6} - 2x + k2 pi    x -  dfrac{ pi }{4} =  pi  -  left( { -  dfrac{{5 pi }}{6} - 2x}  right) + k2 pi   end{array}  right.     Leftrightarrow  left[  begin{array}{l} 3x =  -  dfrac{{7 pi }}{{12}} + k2 pi    x -  dfrac{ pi }{4} =  dfrac{{11 pi }}{6} + 2x + k2 pi   end{array}  right.  Leftrightarrow  left[  begin{array}{l} x =  -  dfrac{{7 pi }}{{36}} +  dfrac{{k2 pi }}{3}   x =  -  dfrac{{25 pi }}{{12}} - k2 pi   end{array}  right. left( {k  in  mathbb{Z}}  right)  end{array}$