Giải phương trình: sin (x/2 + pi/4) = - căn bậc hai 2 / 2
Giải phương trình:
\(\cos \left( {2x - \frac{\pi }{3}} \right) = \sin \left( {\frac{\pi }{4} - x} \right)\);
\(\cos \left( {2x - \frac{\pi }{3}} \right) = \sin \left( {\frac{\pi }{4} - x} \right)\)
\( \Leftrightarrow \cos \left( {2x - \frac{\pi }{3}} \right) = \cos \left[ {\frac{\pi }{2} - \left( {\frac{\pi }{4} - x} \right)} \right]\)
\( \Leftrightarrow \cos \left( {2x - \frac{\pi }{3}} \right) = \cos \left( {\frac{\pi }{4} + x} \right)\)
\( \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{3} = \frac{\pi }{4} + x + k2\pi \\2x - \frac{\pi }{3} = - \left( {\frac{\pi }{4} + x} \right) + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)
\( \Leftrightarrow \left[ \begin{array}{l}2x - x = \frac{\pi }{4} + \frac{\pi }{3} + k2\pi \\2x + x = - \frac{\pi }{4} + \frac{\pi }{3} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)
\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{{7\pi }}{{12}} + k2\pi \\3x = \frac{\pi }{{12}} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)
\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{{7\pi }}{{12}} + k2\pi \\x = \frac{\pi }{{36}} + k\frac{{2\pi }}{3}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\).