\(\begin{array}{l}
a)\quad 2\cos2x + \sin x - 1 =0\\
\Leftrightarrow 2(1 - 2\sin^2x) + \sin x - 1 =0\\
\Leftrightarrow 4\sin^2x - \sin x - 1 =0\\
\Leftrightarrow \left[\begin{array}{l}
\sin x = \dfrac{1 - \sqrt{17}}{8}\\
\sin x = \dfrac{1 + \sqrt{17}}{8}
\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}
x = \arcsin\left(\dfrac{1 - \sqrt{17}}{8}\right) + k2\pi\\
x = \pi - \arcsin\left(\dfrac{1 - \sqrt{17}}{8}\right) + k2\pi\\
x = \arcsin\left(\dfrac{1 + \sqrt{17}}{8}\right) + k2\pi\\
x = \pi - \arcsin\left(\dfrac{1 + \sqrt{17}}{8}\right) + k2\pi
\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy phương trình có các họ nghiệm là}\\
x = \arcsin\left(\dfrac{1 - \sqrt{17}}{8}\right) + k2\pi\\
x = \pi - \arcsin\left(\dfrac{1 - \sqrt{17}}{8}\right) + k2\pi\\
x = \arcsin\left(\dfrac{1 + \sqrt{17}}{8}\right) + k2\pi\\
x = \pi - \arcsin\left(\dfrac{1 + \sqrt{17}}{8}\right) + k2\pi\\
\text{với}\ k\in\Bbb Z\\
b)\quad -2\sin x + \cos x + 1 =0\\
\Leftrightarrow -\dfrac{2}{\sqrt5}\sin x + \dfrac{1}{\sqrt5}\cos x = -\dfrac{1}{\sqrt5}\\
Do\ \left(\dfrac{1}{\sqrt5}\right)^2 + \left(-\dfrac{2}{\sqrt5}\right)^2 = 1\\
\text{Đặt}\ \begin{cases}\cos\alpha = \dfrac{1}{\sqrt5}\\\sin\alpha = \dfrac{2}{\sqrt5}\end{cases}\\
\text{Phương trình trở thành:}\\
\quad \cos x.\cos\alpha - \sin x.\sin\alpha = - cos\alpha\\
\Leftrightarrow \cos(x + \alpha) = \cos\left(\pi - \alpha\right)\\
\Leftrightarrow \left[\begin{array}{l}x + \alpha = \pi -\alpha + k2\pi\\x - \alpha =- \pi +\alpha + \dfrac{\pi}{2} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \pi - 2\alpha + k\pi\\x = -\pi + 2\alpha + k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy phương trình có họ nghiệm là}\\
x = \pi - 2\alpha + k\pi\\
x = -\pi + 2\alpha + k2\pi\\
\text{với $\alpha$ thỏa mãn}\ \begin{cases}\cos\alpha = \dfrac{1}{\sqrt5}\\\sin\alpha = \dfrac{2}{\sqrt5}\end{cases}\ \text{và}\ k\in\Bbb Z
\end{array}\)
