Đáp án: $A$
Giải thích các bước giải:
Ta có:
$2\cos2x+2\cos x-\sqrt{2}=0$
$\to 2(2\cos^2x-1)+2\cos x-\sqrt{2}=0$
$\to 4(\cos^2x-\dfrac12)+2(\cos x-\dfrac{1}{\sqrt{2}})=0$
$\to 4(\cos x-\dfrac{1}{\sqrt{2}})(\cos x+\dfrac{1}{\sqrt{2}})+2(\cos x-\dfrac{1}{\sqrt{2}})=0$
$\to 2(\cos x-\dfrac{1}{\sqrt{2}})(\cos x+\dfrac{1}{\sqrt{2}})+(\cos x-\dfrac{1}{\sqrt{2}})=0$
$\to (\cos x-\dfrac{1}{\sqrt{2}})(2\cos x+\sqrt{2}+1)=0$
Mà $2\cos x+\sqrt{2}+1\ge 2\cdot (-1)+\sqrt{2}+1>0$
$\to \cos x-\dfrac{1}{\sqrt{2}}=0$
$\to \cos x=\dfrac{1}{\sqrt{2}}$
$\to x=\pm\dfrac{\pi}{4}+k2\pi$
Mà $x\in[0,3\pi]$
$\to x\in\{\dfrac{\pi}{4}, \dfrac{9\pi}{4},\dfrac{7\pi}{4}\}$
$\to T=\dfrac{17\pi}{4}$
