sin^6x+cos^6x=1/4sin2x

Đáp án:

\[x = \dfrac{\pi }{4} + k\pi \]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
{\sin ^6}x + {\cos ^6}x = \dfrac{1}{4}\sin 2x\\
 \Leftrightarrow {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3} - 3{\sin ^2}x.{\cos ^2}x.\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = \dfrac{1}{4}\sin 2x\\
 \Leftrightarrow {1^3} - 3.{\sin ^2}x.{\cos ^2}x.1 = \dfrac{1}{4}\sin 2x\\
 \Leftrightarrow 1 - \dfrac{3}{4}.{\left( {2\sin x.\cos x} \right)^2} = \dfrac{1}{4}\sin 2x\\
 \Leftrightarrow 1 - \dfrac{3}{4}{\sin ^2}2x = \dfrac{1}{4}\sin 2x\\
 \Leftrightarrow 4 - 3{\sin ^2}2x = \sin 2x\\
 \Leftrightarrow 3{\sin ^2}2x + \sin 2x - 4 = 0\\
 \Leftrightarrow \left( {\sin 2x - 1} \right)\left( {3\sin 2x + 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 1\\
\sin 2x =  - \dfrac{4}{3}
\end{array} \right.\\
 - 1 \le \sin 2x \le 1 \Rightarrow \sin 2x = 1 \Rightarrow 2x = \dfrac{\pi }{2} + k2\pi  \Leftrightarrow x = \dfrac{\pi }{4} + k\pi 
\end{array}\)