Giải pt. 1) tan(x – bi/3)=cot2x 2) cos^2x= 1 + sin7x

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1. Đáp án:

   \(1)\,\,x = \frac{{5\pi }}{{18}} + \frac{{l\pi }}{3}\,\,\,\left( {l \in Z} \right)\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   1)\,\,\tan \left( {x – \frac{\pi }{3}} \right) = \cot \left( {2x} \right)\,\,\,\,\left( * \right)\\
   DK:\,\,\,\left\{ \begin{array}{l}
   \cos \left( {x – \frac{\pi }{3}} \right) \ne 0\\
   \sin \left( {2x} \right) \ne 0
   \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
   x – \frac{\pi }{3} \ne \frac{\pi }{2} + k\pi \\
   2x \ne m\pi 
   \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
   x \ne \frac{{5\pi }}{6} + k\pi \\
   x \ne \frac{{m\pi }}{2}
   \end{array} \right.\,\,\,\left( {k,\,\,m \in Z} \right).\\
   \left( * \right) \Leftrightarrow \tan \left( {x – \frac{\pi }{3}} \right) = \tan \left( {\frac{\pi }{2} – 2x} \right)\\
    \Leftrightarrow x – \frac{\pi }{3} = \frac{\pi }{2} – 2x + l\pi \\
    \Leftrightarrow x = \frac{{5\pi }}{{18}} + \frac{{l\pi }}{3}\,\,\,\left( {l \in Z} \right)\\
    \Rightarrow pt\,\,co\,\,nghiem \Leftrightarrow \left\{ \begin{array}{l}
   \frac{{5\pi }}{{18}} + \frac{{l\pi }}{3} \ne \frac{{5\pi }}{6} + k\pi \\
   \frac{{5\pi }}{{18}} + \frac{{l\pi }}{3} \ne \frac{{m\pi }}{2}
   \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
   l \ne \frac{5}{3} + 3k\\
   l \ne  – \frac{5}{6} + \frac{{3m}}{2}
   \end{array} \right.\,\,\,tm\,\,\forall m,\,\,l,\,\,k \in Z.
   \end{array}\)

   Đề bài câu 2 của bạn là: \({\cos ^2}x = 1 + \sin 7x\) à bạn?

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