# Câu 35 trang 42 SGK Đại số và Giải tích 11 Nâng cao

##### Hướng dẫn giải

a.

\eqalign{& {\sin ^2}4x + {\sin ^2}3x = {\sin ^2}2x + {\sin ^2}x \cr & \Leftrightarrow {1 \over 2}\left( {1 - \cos 8x} \right) + {1 \over 2}\left( {1 - \cos 6x} \right) = {1 \over 2}\left( {1 - \cos 4x} \right) + {1 \over2}\left( {1 - \cos 2x} \right) \cr & \Leftrightarrow \cos 8x + \cos 6x = \cos 4x + \cos 2x \cr & \Leftrightarrow \cos 7x\cos x = \cos 3x\cos x \cr & \Leftrightarrow \cos x\left( {\cos 7x - \cos 3x} \right) = 0 \cr & \Leftrightarrow \left[ {\matrix{{\cos x = 0} \cr {\cos 7x = \cos 3x} \cr} } \right. \Leftrightarrow \left[ {\matrix{{x = {\pi \over 2} + k\pi } \cr {x = k{\pi \over 2}} \cr {x = k{\pi \over 5}} \cr} } \right. \Leftrightarrow \left[ {\matrix{{x = k{\pi \over 2}} \cr {x = k{\pi \over 5}} \cr} } \right.\,\,\,k \in\mathbb Z \cr}

b. Ta có:

\eqalign{& {\cos ^2}x + {\cos ^2}2x + {\cos ^2}3x + {\cos ^2}4x = 2 \cr & \Leftrightarrow {{1 + \cos 2x} \over 2} + {{1 + \cos 4x} \over 2} + {{1 + \cos 6x} \over 2} + {{1 + \cos 8x} \over 2} = 2 \cr & \Leftrightarrow \left( {\cos 2x + \cos 4x} \right) + \left( {\cos 6x + \cos 8x} \right) = 0 \cr & \Leftrightarrow 2\cos 3x\cos x + 2\cos 7x\cos x = 0 \cr & \Leftrightarrow \cos x\left( {\cos 3x + \cos 7x} \right) = 0 \cr & \Leftrightarrow 2\cos x\cos 5x\cos 2x = 0 \Leftrightarrow \left[ {\matrix{{\cos x = 0} \cr {\cos 2x = 0} \cr {\cos 5x = 0} \cr} } \right. \Leftrightarrow \left[ {\matrix{{x = {\pi \over 2} + k\pi } \cr {x = {\pi \over 4} + k{\pi \over 2}} \cr {x = {\pi \over {10}} + k{\pi \over 5}} \cr} } \right.\,\,\left( {k \in\mathbb Z} \right) \cr}