Giải pt: sin^2(4x) + sin^2(3x) = sin^2(2x) + sin^2(x)

Giải thích các bước giải:

Áp dụng:

\[\begin{array}{l}
\sin a - \sin b = 2.\cos \left( {\frac{{a + b}}{2}} \right).\sin \left( {\frac{{a - b}}{2}} \right)\\
\sin a + \sin b = 2\sin \left( {\frac{{a + b}}{2}} \right).\cos \left( {\frac{{a - b}}{2}} \right)
\end{array}\]

 Ta có:

\(\begin{array}{l}
{\sin ^2}4x + {\sin ^2}3x = {\sin ^2}2x + {\sin ^2}x\\
 \Leftrightarrow \left( {{{\sin }^2}4x - {{\sin }^2}2x} \right) + \left( {{{\sin }^2}3x - {{\sin }^2}x} \right) = 0\\
 \Leftrightarrow \left( {\sin 4x - \sin 2x} \right)\left( {\sin 4x + \sin 2x} \right) + \left( {\sin 3x - {\mathop{\rm sinx}\nolimits} } \right)\left( {\sin 3x + \sin x} \right) = 0\\
 \Leftrightarrow 2.\cos 3x.\sin x.2\sin 3x.\cos x + 2\cos 2x.sinx.2sin2x.cosx = 0\\
 \Leftrightarrow \left( {\sin x.\cos x} \right).\left( {\cos 3x.\sin 3x} \right) + \left( {\cos 2x.sin2x} \right).\left( {\sin x.\cos x} \right) = 0\\
 \Leftrightarrow \frac{1}{2}\sin 2x.\frac{1}{2}.\sin 6x + \frac{1}{2}\sin 2x.\frac{1}{2}sin4x = 0\\
 \Leftrightarrow \sin 2x\left( {\sin 6x + \sin 4x} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\sin 6x =  - \sin 4x
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\sin 6x = \sin \left( { - 4x} \right)
\end{array} \right.
\end{array}\)