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Phương trình đã cho tương đương với:
$\sin x+\cos x=\sqrt2 \sin 5x$
$\Leftrightarrow \dfrac{1}{\sqrt2}\sin x+\dfrac{1}{\sqrt2}\cos x=\sin5x$
$\Leftrightarrow \sin(x+\dfrac{\pi}{4})=\sin5x$
$\Leftrightarrow \left[\begin{array}{l}x+\dfrac{\pi}{4}=5x+k2\pi\\x+\dfrac{\pi}{4}=\pi-5x+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{\pi}{16}-k\dfrac{\pi}{2}\\x=\dfrac{\pi}{8}+k\dfrac{\pi}{3}\end{array}\right.,k\in\mathbb{Z}$
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