Giải phương trình: 1 + sin ^32x + cos ^32x = 1/2sin 4x

Trả lời

Lời giải:

\(1 + {\sin ^3}2x + {\cos ^3}2x = \frac{1}{2}\sin 4x\)

\( \Leftrightarrow 1 + \left( {\sin 2x + \cos 2x} \right)\left( {1 - \sin 2x\cos 2x} \right) = \sin 2x\)

\( \Leftrightarrow \left( {1 + \sin 2x + \cos 2x} \right)\left( {1 - \sin 2x\cos 2x} \right) = 0\)

\[ \Leftrightarrow \sin \left( {2x + \frac{\pi }{4}} \right) = \sin \left( { - \frac{\pi }{4}} \right)\]

\[ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x + \frac{\pi }{4} = - \frac{\pi }{4} + k2\pi }\\{2x + \frac{\pi }{4} = \pi + \frac{\pi }{4} + k2\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right)\]

\[ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = - \frac{\pi }{4} + k\pi }\\{x = \frac{\pi }{2} + k\pi }\end{array}\left( {k \in \mathbb{Z}} \right)} \right.\].