Câu 34 trang 42 SGK Đại số và Giải tích 11 Nâng cao

\(\cos x\cos 5x = \cos 2x\cos 4x\)

Lời giải chi tiết:

Ta có:

\(\eqalign{& \cos x\cos 5x = \cos 2x\cos 4x \cr & \Leftrightarrow {1 \over 2}\left( {\cos 6x + \cos 4x} \right) = {1 \over 2}\left( {\cos 6x + \cos 2x} \right)\cr& \Leftrightarrow \cos 6x + \cos 4x = \cos 6x + \cos 2x\cr&\Leftrightarrow \cos 4x = \cos 2x \cr & \Leftrightarrow \left[ {\matrix{{4x = 2x + k2\pi } \cr {4x = - 2x + k2\pi } \cr} } \right.\cr& \Leftrightarrow \left[ {\matrix{{x = k\pi } \cr {x = k{\pi \over 3}} \cr} } \right. \cr&\Leftrightarrow x = k{\pi \over 3} \,\,(k\in\mathbb Z)\cr} \)

\(\cos 5x\sin 4x=\cos 3x\sin 2x\)

Lời giải chi tiết:

\(\eqalign{& \cos 5x\sin 4x = \cos 3x\sin 2x \cr&\Leftrightarrow {1 \over 2}\left( {\sin 9x - \sin x} \right) = {1 \over 2}\left( {\sin 5x - \sin x} \right) \cr & \Leftrightarrow \sin 9x - \sin x = \sin 5x - \sin x\cr&\Leftrightarrow \sin 9x = \sin 5x \cr&\Leftrightarrow \left[ {\matrix{{9x = 5x + k2\pi } \cr {9x = \pi - 5x + k2\pi } \cr} } \right. \cr&\Leftrightarrow \left[ {\matrix{{x = k{\pi \over 2}} \cr {x = {\pi \over {14}} + k{\pi \over 7}} \cr} } \,\,(k\in\mathbb Z) \right. \cr} \)

\(\sin 2x + \sin 4x = \sin 6x\)

Lời giải chi tiết:

\(\eqalign{& \sin 2x + \sin 4x = \sin 6x \cr&\Leftrightarrow 2\sin 3x\cos x = 2\sin 3x\cos 3x \cr & \Leftrightarrow \sin 3x\left( {\cos x - \cos 3x} \right) = 0 \cr&\Leftrightarrow \left[ {\matrix{{\sin 3x = 0} \cr {\cos x = \cos 3x} \cr} } \right.\cr&\Leftrightarrow \left[ \begin{array}{l}3x = k\pi \\3x = x + k2\pi \\3x = - x + k2\pi \end{array} \right.\cr&\Leftrightarrow \left[ {\matrix{{x = k{\pi \over 3}} \cr {x = k\pi } \cr {x = k{\pi \over 2}} \cr} } \right. \cr&\Leftrightarrow \left[ {\matrix{{x = k{\pi \over 3}} \cr {x = k{\pi \over 2}} \cr} } \,\,(k\in\mathbb Z)\right. \cr} \)

\(\sin x + \sin 2x = \cos x + \cos 2x\)

Lời giải chi tiết:

\(\eqalign{& \sin x + \sin 2x = \cos x + \cos 2x \cr&\Leftrightarrow 2\sin {{3x} \over 2}\cos {x \over 2} = 2\cos {{3x} \over 2}\cos {x \over 2} \cr & \Leftrightarrow \cos {x \over 2}\left( {\sin {{3x} \over 2} - \cos {{3x} \over 2}} \right) = 0 \cr&\Leftrightarrow \left[ {\matrix{{\cos {x \over 2} = 0} \cr {\sin {{3x} \over 2} = \cos {{3x} \over 2}} \cr} } \right. \cr & \Leftrightarrow \left[ {\matrix{{{x \over 2} = {\pi \over 2} + k\pi } \cr {\tan {{3x} \over 2} = 1} \cr} } \right.\cr& \Leftrightarrow \left[ {\matrix{{x = \pi + k2\pi } \cr {x = {\pi \over 6} + k{{2\pi } \over 3}} \cr} } \right.\left( {k \in\mathbb Z} \right) \cr} \)