Giải phương trình \(f’\left( x \right) = g\left( x \right),\) biết rằng
a) \(f\left( x \right) = {{1 - \cos 3x} \over 3};g\left( x \right) = \left( {\cos 6x - 1} \right)\cot 3x.\)
b) \(f\left( x \right) = {1 \over 2}\cos 2x;g\left( x \right) = 1 - {\left( {\cos 3x + \sin 3x} \right)^2}.\)
c) \(f\left( x \right) = {1 \over 2}\sin 2x + 5\cos x;g\left( x \right) = 3{\sin ^2}x + {3 \over {1 + {{\tan }^2}x}}.\)
a) \(f\left( x \right) = {{1 - \cos 3x} \over 3} \Rightarrow f’\left( x \right) = \sin 3x.\) Ta có
\(f’\left( x \right) = g\left( x \right) \Leftrightarrow \left( {\cos 6x - 1} \right).\cot 3x = \sin 3x\) (điều kiện: \(\sin 3x \ne 0 \Leftrightarrow \cos 3x \ne \pm 1\) )
\(\eqalign{
& \Leftrightarrow \left( {\cos 6x - 1} \right).\cos 3x = {\sin ^2}3x \cr
& \Leftrightarrow \left( {1 - 2{{\sin }^2}3x - 1} \right).\cos 3x = {\sin ^2}3x \cr
& \Leftrightarrow {\sin ^2}3x.\left( {2\cos 3x + 1} \right) = 0 \cr
& \Leftrightarrow \cos 3x = - {1 \over 2}{\rm{ }}\left( {{\rm{vì}}\,\,\sin 3x \ne 0{\rm{ }}} \right) \cr
& \Leftrightarrow \cos 3x = \cos {{2\pi } \over 3} \cr
& \Leftrightarrow 3x = \pm {{2\pi } \over 3} + k2\pi \cr
& \Leftrightarrow x = \pm {{2\pi } \over 9} + k{{2\pi } \over 3}{\rm{ }}\left( {k \in Z} \right). \cr} \)
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b) \(f\left( x \right) = {1 \over 2}\cos 2x \Rightarrow f’\left( x \right) = - \sin 2x.\) Ta có
\(\eqalign{
& f’\left( x \right) = g\left( x \right) \cr
& \Leftrightarrow - \sin 2x = 1 - {\left( {\cos 3x + \sin 3x} \right)^2} \cr
& \Leftrightarrow 1 + \sin 2x = {\left( {\cos 3x + \sin 3x} \right)^2} \cr
& \Leftrightarrow 1 + \sin 2x = 1 + 2\sin 3x\cos 3x \cr
& \Leftrightarrow \sin 6x - \sin 2x = 0 \cr
& \Leftrightarrow 2\cos 4x\sin 2x = 0 \cr
& \Leftrightarrow \left[ \matrix{
\cos 4x = 0 \hfill \cr
\sin 2x = 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
4x = {\pi \over 2} + k\pi \hfill \cr
2x = n\pi \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = {\pi \over 8} + k{\pi \over 4} \hfill \cr
x = n{\pi \over 2} \hfill \cr} \right.\left( {k,n \in Z} \right). \cr}\)
c) \(f\left( x \right) = {1 \over 2}\sin 2x + 5\cos x \Rightarrow f’\left( x \right) = \cos 2x - 5\sin x.\) Ta có
\(\eqalign{
& f’\left( x \right) = g\left( x \right) \cr
& \Leftrightarrow \cos 2x - 5\sin x = 3{\sin ^2}x + {3 \over {1 + {{\tan }^2}x}} \cr
& \Leftrightarrow 5\sin x + {3 \over {1 + {{\tan }^2}x}} = \cos 2x - 3{\sin ^2}x \cr
& \Leftrightarrow 5\sin x + 3{\cos ^2}x = {\cos ^2}x - 4{\sin ^2}x \cr
& \Leftrightarrow 5\sin x = - 2{\cos ^2}x - 4{\sin ^2}x \cr
& \Leftrightarrow 5\sin x = - 2 - 2{\sin ^2}x \cr
& \Leftrightarrow 2{\sin ^2}x + 5\sin x + 2 = 0. \cr} \)
Đặt \(t = \sin x,t \in \left[ { - 1;1} \right],\) ta có phương trình \(2{t^2} + 5t + 2 = 0.\)
Giải phương trình \(t = - {1 \over 2}\) ta được (loại t = -2 ).
\(\eqalign{
& \sin x = - {1 \over 2} \cr
& \Leftrightarrow \sin x = \sin \left( { - {\pi \over 6}} \right) \cr
& \Leftrightarrow \left[ \matrix{
x = - {\pi \over 6} + k2\pi \hfill \cr
x = {{7\pi } \over 6} + k2\pi \hfill \cr} \right.\left( {k \in Z} \right). \cr} \)