$\begin{array}{l}a)\,\,2\sin^2x – 5\sin x + 2 = 0\\ \Leftrightarrow (\sin x – 2)\left(\sin x – \dfrac{1}{2}\right) =0\\ \Leftrightarrow \left[\begin{array}{l}\sin x = 2\qquad (loại)\\\sin x = \dfrac{1}{2}\quad (nhận)\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} +k2\pi\\x = \dfrac{5\pi}{6} +k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\ b) \,\,\sin2x + \sqrt3\cos2x= 1\\ \Leftrightarrow \dfrac{1}{2}\sin2x + \dfrac{\sqrt3}{2}\cos2x = \dfrac{1}{2}\\ \Leftrightarrow \sin\left(2x + \dfrac{\pi}{3}\right) = \sin\dfrac{\pi}{6}\\ \Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{3} = \dfrac{\pi}{6} + k2\pi\\2x + \dfrac{\pi}{3} = \dfrac{5\pi}{6} +k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{12} + k\pi\\x = \dfrac{\pi}{4} +k\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$ Reply
`a) 2sin² x – 5sin x + 2 = 0` `<=>` \(\left[ \begin{array}{l}sin x = 2 (l)\\sin x = \dfrac{1}{2}\end{array} \right.\) `=> sin x = 1/2` `<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{6} + k2π\\x = \dfrac{5π}{6} + k2π\end{array} \right.\) `(k ∈ ZZ)` `b) sin 2x + sqrt{3}cos 2x = 1` `<=> 1/(2)sin 2x + (\sqrt{3})/(2)cos 2x = 1/2` `<=> sin (2x + π/3) = sin (π/6)` `<=>` \(\left[ \begin{array}{l}2x + \dfrac{π}{3} = \dfrac{π}{6} + k2π\\2x + \dfrac{π}{3} = \dfrac{5π}{6} + k2π\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x = -\dfrac{π}{12} + kπ\\x = \dfrac{π}{4} + kπ\end{array} \right.\) `(k ∈ ZZ)` Reply
$\begin{array}{l}a)\,\,2\sin^2x – 5\sin x + 2 = 0\\ \Leftrightarrow (\sin x – 2)\left(\sin x – \dfrac{1}{2}\right) =0\\ \Leftrightarrow \left[\begin{array}{l}\sin x = 2\qquad (loại)\\\sin x = \dfrac{1}{2}\quad (nhận)\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} +k2\pi\\x = \dfrac{5\pi}{6} +k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\ b) \,\,\sin2x + \sqrt3\cos2x= 1\\ \Leftrightarrow \dfrac{1}{2}\sin2x + \dfrac{\sqrt3}{2}\cos2x = \dfrac{1}{2}\\ \Leftrightarrow \sin\left(2x + \dfrac{\pi}{3}\right) = \sin\dfrac{\pi}{6}\\ \Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{3} = \dfrac{\pi}{6} + k2\pi\\2x + \dfrac{\pi}{3} = \dfrac{5\pi}{6} +k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{12} + k\pi\\x = \dfrac{\pi}{4} +k\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
`a) 2sin² x – 5sin x + 2 = 0`
`<=>` \(\left[ \begin{array}{l}sin x = 2 (l)\\sin x = \dfrac{1}{2}\end{array} \right.\)
`=> sin x = 1/2`
`<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{6} + k2π\\x = \dfrac{5π}{6} + k2π\end{array} \right.\) `(k ∈ ZZ)`
`b) sin 2x + sqrt{3}cos 2x = 1`
`<=> 1/(2)sin 2x + (\sqrt{3})/(2)cos 2x = 1/2`
`<=> sin (2x + π/3) = sin (π/6)`
`<=>` \(\left[ \begin{array}{l}2x + \dfrac{π}{3} = \dfrac{π}{6} + k2π\\2x + \dfrac{π}{3} = \dfrac{5π}{6} + k2π\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = -\dfrac{π}{12} + kπ\\x = \dfrac{π}{4} + kπ\end{array} \right.\) `(k ∈ ZZ)`