3sin^2x +sin2x+cos^2x = 3 ( kí hiệu: ^ là mũ nha)

$3{\sin}^2x +\sin2x+{\cos}^2x = 3 $ $\Rightarrow 3\dfrac{1-\cos 2x}{2}+\sin 2x+\dfrac{\cos 2x+1}{2}=3$ $\Rightarrow -\cos 2x+\sin 2x=1$ $\Rightarrow \sqrt2\sin \left({2x-\dfrac{\pi}{4}}\right)=1$ $\Rightarrow \sin \left({2x-\dfrac{\pi}{4}}\right)=\dfrac{1}{\sqrt2} $ $\Rightarrow \left[ \begin{array}{l} 2x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi \\ 2x-\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi \end{array} \right .$ $\Rightarrow \left[ \begin{array}{l} x=\dfrac{\pi}{2}+k\pi \\ x=\dfrac{\pi}{2}+k\pi \end{array} \right .(k\in\mathbb Z)$