Giải thích các bước giải:
a.Ta có :
$\sin 2x+\sqrt{2}\cos 2x=1$
$\to\dfrac{\sqrt{2}}{\sqrt{3}}\cos 2x+\dfrac{1}{\sqrt{3}}\sin 2x=1$
Đặt $\alpha =\arccos \dfrac{\sqrt{2}}{\sqrt{3}}$
$\rightarrow \cos \alpha\cos 2x+\sin \alpha\sin 2x=1$
$\rightarrow \cos(2x -\alpha)=1$
$\rightarrow 2x -\alpha=k\pi$
$\rightarrow x =\dfrac{\alpha+k\pi}{2}$
b.$4\sin 6x+\sin 10x-2\sin 2x.\cos 4x=0$
$\rightarrow 4\sin 6x+\sin 10x-(\sin 6x+\sin -2x)=0$
$\rightarrow 4\sin 6x+\sin 10x-(\sin 6x-\sin 2x)=0$
$\rightarrow 3\sin 6x+\sin 10x+\sin 2x=0$
