1. sin5x-cos(x + 5pi/2) = sin(2x – pi/2)

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1. Đáp án:

   $\begin{array}{l}
   \sin 5x – \cos \left( {x + \dfrac{{5\pi }}{2}} \right) = \sin \left( {2x – \dfrac{\pi }{2}} \right)\\
    \Rightarrow \sin 5x + \sin x =  – \cos 2x\\
    \Rightarrow 2.\sin \dfrac{{5x + x}}{2}.\cos \dfrac{{5x – x}}{2} =  – \cos 2x\\
    \Rightarrow 2.sin3x.cos2x + cos2x = 0\\
    \Rightarrow cos2x.\left( {2\sin 3x + 1} \right) = 0\\
    \Rightarrow \left[ \begin{array}{l}
   \cos 2x = 0\\
   2\sin 3x + 1 = 0
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   2x = \dfrac{\pi }{2} + k\pi \\
   \sin 3x =  – \dfrac{1}{2}
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
   3x = \dfrac{{ – \pi }}{6} + k2\pi \\
   3x = \dfrac{{7\pi }}{6} + k2\pi 
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
   x = \dfrac{{ – \pi }}{{18}} + \dfrac{{k2\pi }}{3}\\
   x = \dfrac{{7\pi }}{{18}} + \dfrac{{k2\pi }}{3}
   \end{array} \right.
   \end{array}$

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