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trung70811av

$sin(4x+\frac{\pi}{4})sin6x=sin(10x+\frac{\pi}{4})$ (1)

Giải:
pt (1) \Leftrightarrow $(sin4x+cos4x)sin6x=sin4xcos6x+cos4xsin6x +sin6xcos4x +sin6xsin4x-cos4xcos6x$
\Leftrightarrow $sin4xcos6x-cos4xcos6x=0$
\Leftrightarrow $cos6x=0$ \bigcup_{}^{} $sin4x-cos4x=0$
\Leftrightarrow $cos(6x)=0$ \bigcup_{}^{} $cos(4x+\frac{\pi}{4})=0$
\Leftrightarrow $x=\frac{\pi}{12}+\frac{k\pi}{3}$ \bigcup_{}^{} $x=\frac{-\pi}{12}+\frac{k\pi}{3}$ \bigcup_{}^{} $x=\frac{\pi}{16}+\frac{k\pi}{2}$ \bigcup_{}^{} $x=\frac{-3\pi}{16}+\frac{k\pi}{2}$
 
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