Sin^4x+cos^4x=1/2sin2x

Đáp án:

\[x = \dfrac{\pi }{4} + k\pi \,\,\,\,\left( {k \in Z} \right)\]

Giải thích các bước giải:

 Ta có:

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