Đáp án:
$\left\{\begin{array}{l} \dfrac{\pi}{6} + 2k\pi\\x = \dfrac{5\pi}{6} + 2k\pi\end{array}\right. \,\, (k \in Z)$
Giải thích các bước giải:
$\sin2x-\cos2x+3\sin x-\cos x-1=0$
$\Leftrightarrow 2\sin{x}.\cos{x} - 1 + 2\sin^2{x} + 3\sin{x} - \cos{x} - 1 = 0$
$\Leftrightarrow \cos{x}(2\sin{x} - 1) + \sin{x}(2\sin{x} - 1) + 2(2\sin{x} - 1) = 0$
$\Leftrightarrow (2\sin{x} - 1)(\cos{x} + \sin{x} + 2) = 0$
$\Leftrightarrow \left[\begin{array}{l} \sin{x} = \dfrac{1}{2}\\\sin{x} + \cos{x} = -2\end{array}\right. \,\, (1)$
Ta thấy:
$\sin{x} + \cos{x} = \sqrt{2}.\sin{(x + \dfrac{\pi}{4})} \in [- \sqrt{2}; \sqrt{2}] \Rightarrow \sin{x} + \cos{x} > -2$
Do đó:
$(1) \Leftrightarrow \sin{x} = \dfrac{1}{2} \Rightarrow \left[\begin{array}{l} \dfrac{\pi}{6} + 2k\pi\\x = \dfrac{5\pi}{6} + 2k\pi\end{array}\right. \,\, (k \in Z)$
