Giải phương trình cos ^3x + sin ^3x = cos 2x

Trả lời

Lời giải:

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

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

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

\(\begin{array}{l} \Leftrightarrow \left( {\cos x + \sin x} \right)\left( {1 - \sin x\cos x - \cos x + \sin x} \right) = 0\\ \Leftrightarrow \left( {\sin x + \cos x} \right)\left[ {\left( {1 + \sin x} \right) - \cos x\left( {\sin x + 1} \right)} \right] = 0\end{array}\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sin x = - \cos x}\\{\sin x = - 1}\\{\cos x = 1}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\tan x = - 1}\\{\sin x = - 1}\\{\cos x = 1}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = - \frac{\pi }{4} + k\pi }\\{x = - \frac{\pi }{2} + k2\pi }\\{x = k\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right)\).