Toán Lớp 11: giải phương trình sin^2(3x)-cos^2(4x)=sin^2(5x)-cos^(6x) giải giúp m vs mn ơi

Question

Toán Lớp 11:  giải phương trình sin^2(3x)-cos^2(4x)=sin^2(5x)-cos^(6x) giải giúp m vs mn ơi

dinhcongson · Answer

sin^2 3x-cos^2 4x=sin^2 5x-cos^2 6x   ->(1-cos6x)/2-(1+cos8x)/2=(1-cos10x)/2-(1+cos12x)/2   ->cos12x-cos6x+cos10x-cos8x=0   ->-2sin9xsin3x-2sin9xsinx=0   ->sin9x(sin3x+sinx)=0   ->  ( left[  begin{array}{l}sin9x=0  sin3x=-sinx=sin(-x) end{array}  right. )   -> ( left[  begin{array}{l}x=k pi/9  x=k pi/2, pi/2+k pi(k&isin;Z) end{array}  right. )

tongtung · Answer

$ begin{array}{l} { sin ^2}3x - { cos ^2}4x = { sin ^2}5x - { cos ^2}6x     Leftrightarrow  dfrac{{1 -  cos 6x}}{2} -  dfrac{{1 +  cos 8x}}{2} =  dfrac{{1 -  cos 10x}}{2} -  dfrac{{1 +  cos 12x}}{2}     Leftrightarrow  -  cos 6x -  cos 8x =  -  cos 10x -  cos 12x     Leftrightarrow  cos 6x +  cos 8x =  cos 10x +  cos 12x     Leftrightarrow 2 cos 7x cos x = 2 cos 11x cos x     Leftrightarrow 2 cos x left( { cos 11x -  cos 7x}  right) = 0     Leftrightarrow  left[  begin{array}{l}  cos x = 0    cos 11x =  cos 7x  end{array}  right.  Leftrightarrow  left[  begin{array}{l} x =  dfrac{ pi }{2} + k pi    11x = 7x + k2 pi    11x =  - 7x + k2 pi   end{array}  right.  Leftrightarrow  left[  begin{array}{l} x =  dfrac{ pi }{2} + k pi    x =  dfrac{{k pi }}{2}   x =  dfrac{{k pi }}{9}  end{array}  right.     Rightarrow  left[  begin{array}{l} x =  dfrac{{k pi }}{2}   x =  dfrac{{k pi }}{9}  end{array}  right. left( {k  in  mathbb{Z}}  right)  end{array}$