Giải phương trình: \( {{\sin }^{2}}x+{{\sin }^{2}}3x=co{{s}^{2}}2x+{{\cos }^{2}}4x \) (*)
Hướng dẫn giải:
Ta có: (*) \( \Leftrightarrow \frac{1}{2}(1-\cos 2x)+\frac{1}{2}(1-\cos 6x)=\frac{1}{2}(1+\cos 4x)+\frac{1}{2}(1+\cos 8x) \)
\(\Leftrightarrow -(\cos 2x+\cos 6x)=\cos 4x+\cos 8x\Leftrightarrow -2\cos 4x\cos 2x=2\cos 6x\cos 2x\)
\(\Leftrightarrow 2\cos 2x(\cos 6x+\cos 4x)=0\Leftrightarrow 4\cos 2x\cos 5x\cos x=0\Leftrightarrow \left[ \begin{align} & \cos 2x=0 \\ & \cos 5x=0 \\ & \cos x=0 \\ \end{align} \right.\)
\( \Leftrightarrow \left[ \begin{align} & 2x=\frac{\pi }{2}+k\pi \\ & 5x=\frac{\pi }{2}+k\pi \\ & x=\frac{\pi }{2}+k\pi \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align} & x=\frac{\pi }{4}+\frac{k\pi }{2} \\ & x=\frac{\pi }{10}+\frac{k\pi }{5} \\ & x=\frac{\pi }{2}+k\pi \\ \end{align} \right.,\text{ }k\in \mathbb{Z} \).
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