Giải phương trình: a) 2 sin ( 3 x + π/ 4 ) = √ 3 ; b) tan ( 2 x − π/ 4 ) = cot x .

a) \(2\sin \left( {3x + \frac{\pi }{4}} \right) = \sqrt 3 \Leftrightarrow \sin \left( {3x + \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2}\)

\[ \Leftrightarrow \left[ \begin{array}{l}3x + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\3x + \frac{\pi }{4} = \pi - \frac{\pi }{3} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3x = \frac{\pi }{3} - \frac{\pi }{4} + k2\pi \\3x = \pi - \frac{\pi }{3} - \frac{\pi }{4} + k2\pi \end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{56}} + k\frac{{2\pi }}{3}\\x = \frac{{5\pi }}{{56}} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right).\]

Vậy nghiệm của phương trình đã cho là \(x = \frac{\pi }{{56}} + k\frac{{2\pi }}{3};\,\,x = \frac{{5\pi }}{{56}} + k\frac{{2\pi }}{3},\,\,k \in \mathbb{Z}\).

b) Điều kiện \[\left\{ \begin{array}{l}\cos \left( {2x - \frac{\pi }{4}} \right) \ne 0\\\sin x \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2x - \frac{\pi }{4} \ne \frac{\pi }{2} + k\pi \\x \ne l\pi \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne \frac{{3\pi }}{8} + \frac{{k\pi }}{2}\\x \ne l\pi \end{array} \right.\,\,\left( {k;\,l \in \mathbb{Z}} \right)\]

Khi đó \(\tan \left( {2x - \frac{\pi }{4}} \right) = \cot x \Leftrightarrow \tan \left( {2x - \frac{\pi }{4}} \right) = \tan \left( {\frac{\pi }{2} - x} \right)\)

\[ \Leftrightarrow 2x - \frac{\pi }{4} = \frac{\pi }{2} - x + k\pi \Leftrightarrow 3x = \frac{{3\pi }}{4} + k\pi \Leftrightarrow x = \frac{\pi }{4} + \frac{{k\pi }}{3}\,\,(k \in \mathbb{Z})\]

Vậy nghiệm của phương trình đã cho là \[x = \frac{\pi }{4} + \frac{{k\pi }}{3}\,\,(k \in \mathbb{Z})\].