Đáp án: $x=\dfrac12(\arctan(-(\sqrt{2}+1))+k\pi)$
Giải thích các bước giải:
ĐKXĐ: $x\ne k\pi$
Ta có:
$\dfrac{\sin x-\sin3x}{\sqrt{2}\sin x}=\sin2x+\cos2x$
$\to \dfrac{\sin x-(3\sin x-4\sin^3x)}{\sqrt{2}\sin x}=\sin2x+\cos2x$
$\to \dfrac{4\sin^3x-2\sin x}{\sqrt{2}\sin x}=\sin2x+\cos2x$
$\to \dfrac{4\sin^2x-2}{\sqrt{2}}=\sin2x+\cos2x$
$\to \dfrac{2(2\sin^2x-1)}{\sqrt{2}}=\sin2x+\cos2x$
$\to \dfrac{2(-\cos2x)}{\sqrt{2}}=\sin2x+\cos2x$
$\to -\sqrt{2}\cos2x=\sin2x+\cos2x$
$\to -\sqrt{2}\cos2x-\cos2x=\sin2x$
$\to -(\sqrt{2}+1)\cos2x=\sin2x$
Nếu $\sin2x=0\to \cos2x=0$ loại
$\to\sin 2x\ne 0$
$\to \dfrac{\sin2x}{\cos2x}=-(\sqrt{2}+1)$
$\to\tan2x=-(\sqrt{2}+1)$
$\to 2x=\arctan(-(\sqrt{2}+1))+k\pi$
$\to x=\dfrac12(\arctan(-(\sqrt{2}+1))+k\pi)$
