Giải các phương trình sau: Sin(2x + 5π/2) – 3cos(x- 7π/2)= 1 + 2sinx sin( x+ π/6) – cos(2x + π/3)= 2 Cos(x – π/6)= 2sin(x/2 – π/3)

Giải thích các bước giải:

$\begin{array}{l}
a)\sin \left( {2x + \dfrac{{5\pi }}{2}} \right) – 3\cos \left( {x – \dfrac{{7\pi }}{2}} \right) = 1 + 2\sin x\\
 \Leftrightarrow \sin \left( {2x + \dfrac{\pi }{2}} \right) – 3\cos \left( {x + \dfrac{\pi }{2}} \right) = 1 + 2\sin x\\
 \Leftrightarrow \cos 2x – 3\sin \left( { – x} \right) = 1 + 2\sin x\\
 \Leftrightarrow 1 – 2{\sin ^2}x + \sin x = 1\\
 \Leftrightarrow 2{\sin ^2}x – \sin x = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x = \dfrac{1}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
b)\sin \left( {x + \dfrac{\pi }{6}} \right) – \cos \left( {2x + \dfrac{\pi }{3}} \right) = 2\\
 \Leftrightarrow \sin \left( {x + \dfrac{\pi }{6}} \right) – \left( {1 – 2{{\sin }^2}\left( {x + \dfrac{\pi }{6}} \right)} \right) = 2\\
 \Leftrightarrow 2{\sin ^2}\left( {x + \dfrac{\pi }{6}} \right) + \sin \left( {x + \dfrac{\pi }{6}} \right) – 3 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x + \dfrac{\pi }{6}} \right) = 1\\
\sin \left( {x + \dfrac{\pi }{6}} \right) = \dfrac{{ – 3}}{2}\left( l \right)
\end{array} \right.\\
 \Leftrightarrow x + \dfrac{\pi }{6} = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\\
 \Leftrightarrow x = \dfrac{\pi }{3} + k2\pi \left( {k \in Z} \right)
\end{array}$

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