Đề bài

Giải phương trình: a) \(\sin \left( {3x - \frac{\pi }{4}} \right) = \sin \left( {x + \frac{\pi }{6}} \right)\)      b) \(\cos \left( {2x - \frac{\pi }{3}} \right) = \sin \left( {\frac{\pi }{4} - x} \right)\) c*) \({\sin ^2}\left( {x + \frac{\pi }{4}} \right) = {\sin ^2}\left( {2x + \frac{\pi }{2}} \right)\)       d*) \({\cos ^2}\left( {2x + \frac{\pi }{2}} \right) = {\sin ^2}\left( {x + \frac{\pi }{6}} \right)\) e) \(\cos x + \sin x = 0\)                     g) \(\sin x - \sqrt 3 \cos x = 0\)

Lời giải chi tiết

a) Ta có:\(\sin \left( {3x - \frac{\pi }{4}} \right) = \sin \left( {x + \frac{\pi }{6}} \right) \Leftrightarrow \left[ \begin{array}{l}3x - \frac{\pi }{4} = x + \frac{\pi }{6} + k2\pi \\3x - \frac{\pi }{4} = \pi  - \left( {x + \frac{\pi }{6}} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}2x = \frac{{5\pi }}{{12}} + k2\pi \\4x = \frac{{13\pi }}{{12}} + k2\pi \end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{{5\pi }}{{24}} + k\pi \\x = \frac{{13\pi }}{{48}} + k\frac{\pi }{2}\end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\)b) Ta có \(\sin \left( {\frac{\pi }{4} - x} \right) = \cos \left( {\frac{\pi }{2} - \frac{\pi }{4} + x} \right) = \cos \left( {\frac{\pi }{4} + x} \right)\). Phương trình trở thành:\(\cos \left( {2x - \frac{\pi }{3}} \right) = \cos \left( {x + \frac{\pi }{4}} \right) \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{3} = x + \frac{\pi }{4} + k2\pi \\2x - \frac{\pi }{3} =  - \left( {x + \frac{\pi }{4}} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{7\pi }}{{12}} + k2\pi \\3x = \frac{\pi }{{12}} + k2\pi \end{array} \right.\) \( \Leftrightarrow \left[ \begin{array}{l}x = \frac{{7\pi }}{{12}} + k2\pi \\x = \frac{\pi }{{36}} + k\frac{{2\pi }}{3}\end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\)c) Sử dụng công thức hạ bậc, ta có:\({\sin ^2}\left( {x + \frac{\pi }{4}} \right) = \frac{{1 - \cos \left[ {2\left( {x + \frac{\pi }{4}} \right)} \right]}}{2} = \frac{{1 - \cos \left( {2x + \frac{\pi }{2}} \right)}}{2}\),\({\sin ^2}\left( {2x + \frac{\pi }{2}} \right) = \frac{{1 - \cos \left[ {2\left( {2x + \frac{\pi }{2}} \right)} \right]}}{2} = \frac{{1 - \cos \left( {4x + \pi } \right)}}{2}\)Phương trình trở thành:\(\frac{{1 - \cos \left( {2x + \frac{\pi }{2}} \right)}}{2} = \frac{{1 - \cos \left( {4x + \pi } \right)}}{2} \Leftrightarrow \cos \left( {2x + \frac{\pi }{2}} \right) = \cos \left( {4x + \pi } \right)\) \( \Leftrightarrow \left[ \begin{array}{l}2x + \frac{\pi }{2} = 4x + \pi  + k2\pi \\2x + \frac{\pi }{2} =  - \left( {4x + \pi } \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - 2x = \frac{\pi }{2} + k2\pi \\6x =  - \frac{{3\pi }}{2} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{\pi }{4} + k\pi \\x =  - \frac{\pi }{4} + k\frac{\pi }{3}\end{array} \right. \Leftrightarrow x =  - \frac{\pi }{4} + k\frac{\pi }{3}\)\(\left( {k \in \mathbb{Z}} \right)\)d) Sử dụng công thức hạ bậc, ta có:\({\cos ^2}\left( {2x + \frac{\pi }{2}} \right) = \frac{{1 + \cos \left[ {2\left( {2x + \frac{\pi }{2}} \right)} \right]}}{2} = \frac{{1 + \cos \left( {4x + \pi } \right)}}{2}\)\({\sin ^2}\left( {x + \frac{\pi }{6}} \right) = \frac{{1 - \cos \left[ {2\left( {x + \frac{\pi }{6}} \right)} \right]}}{2} = \frac{{1 - \cos \left( {2x + \frac{\pi }{3}} \right)}}{2}\) Phương trình trở thành:\(\frac{{1 + \cos \left( {4x + \pi } \right)}}{2} = \frac{{1 - \cos \left( {2x + \frac{\pi }{3}} \right)}}{2} \Leftrightarrow \cos \left( {4x + \pi } \right) =  - \cos \left( {2x + \frac{\pi }{3}} \right)\)Mặt khác, ta có \( - \cos \left( {2x + \frac{\pi }{3}} \right) = \cos \left( {\pi  + 2x + \frac{\pi }{3}} \right) = \cos \left( {2x + \frac{{4\pi }}{3}} \right)\).Phương trình trở thành:\(\cos \left( {4x + \pi } \right) = \cos \left( {2x + \frac{{4\pi }}{3}} \right) \Leftrightarrow \left[ \begin{array}{l}4x + \pi  = 2x + \frac{{4\pi }}{3} + k2\pi \\4x + \pi  =  - \left( {2x + \frac{{4\pi }}{3}} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{3} + k2\pi \\6x =  - \frac{{7\pi }}{3} + k2\pi \end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k\pi \\x =  - \frac{{7\pi }}{{18}} + k\frac{\pi }{3}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k\pi \\x =  - \frac{\pi }{{18}} + k\frac{\pi }{3}\end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\) e) Ta có \(\frac{1}{{\sqrt 2 }}\left( {\sin x + \cos x} \right) = \sin x\cos \frac{\pi }{4} + \cos x\sin \frac{\pi }{4} = \sin \left( {x + \frac{\pi }{4}} \right)\).Do đó, \(\cos x + \sin x = 0 \Leftrightarrow \frac{1}{{\sqrt 2 }}\left( {\cos x + \sin x} \right) = 0 \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\)\( \Leftrightarrow x + \frac{\pi }{4} = k\pi  \Leftrightarrow x =  - \frac{\pi }{4} + k\pi \)\(\left( {k \in \mathbb{Z}} \right)\)f) Nếu \(\cos x = 0\) thì \(\sin x = 0\). Điều này là vô lí, do \({\sin ^2}x + {\cos ^2}x = 1\).Như vậy \(\cos x \ne 0\). Phương trình trở thành:\(\sin x = \sqrt 3 \cos x \Leftrightarrow \frac{{\sin x}}{{\cos x}} = \sqrt 3  \Leftrightarrow \tan x = \sqrt 3 \)Ta có \(\tan \frac{\pi }{3} = \sqrt 3 \). Phương trình trở thành \(\tan x = \tan \frac{\pi }{3} \Leftrightarrow x = \frac{\pi }{3} + k\pi \)\(\left( {k \in \mathbb{Z}} \right)\)