Bài 1: A/sin^2(2x)=1/2 B/cot^2(x/2)=1/3 C/tan(π/12+12x)=-căn3

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1. Đáp án:

   a. \(\left[ \begin{array}{l}
   x = \dfrac{\pi }{{12}} + k\pi \\
   x = \dfrac{{5\pi }}{{12}} + k\pi \\
   x =  – \dfrac{\pi }{{12}} + k\pi \\
   x = \dfrac{{7\pi }}{{12}} + k\pi 
   \end{array} \right.\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   a.{\sin ^2}\left( {2x} \right) = \dfrac{1}{2}\\
    \to \left[ \begin{array}{l}
   \sin 2x = \dfrac{1}{2}\\
   \sin 2x =  – \dfrac{1}{2}
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   2x = \dfrac{\pi }{6} + k2\pi \\
   2x = \dfrac{{5\pi }}{6} + k2\pi \\
   2x =  – \dfrac{\pi }{6} + k2\pi \\
   2x = \dfrac{{7\pi }}{6} + k2\pi 
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x = \dfrac{\pi }{{12}} + k\pi \\
   x = \dfrac{{5\pi }}{{12}} + k\pi \\
   x =  – \dfrac{\pi }{{12}} + k\pi \\
   x = \dfrac{{7\pi }}{{12}} + k\pi 
   \end{array} \right.\left( {k \in Z} \right)\\
   b.DK:\sin \left( {\dfrac{x}{2}} \right) \ne 0\\
   {\cot ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{1}{3}\\
    \to {\tan ^2}\dfrac{x}{2} = 3\\
    \to \left[ \begin{array}{l}
   \tan \dfrac{x}{2} = \sqrt 3 \\
   \tan \dfrac{x}{2} =  – \sqrt 3 
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   \dfrac{x}{2} = \dfrac{\pi }{3} + k\pi \\
   \dfrac{x}{2} =  – \dfrac{\pi }{3} + k\pi 
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x = \dfrac{{2\pi }}{3} + k2\pi \\
   x =  – \dfrac{{2\pi }}{3} + k2\pi 
   \end{array} \right.\left( {k \in Z} \right)\\
   c.DK:\cos \left( {\dfrac{\pi }{{12}} + 12x} \right) \ne 0\\
   \tan \left( {\dfrac{\pi }{{12}} + 12x} \right) =  – \sqrt 3 \\
    \to \dfrac{\pi }{{12}} + 12x =  – \dfrac{\pi }{3} + k\pi \\
    \to 12x =  – \dfrac{{5\pi }}{{12}} + k\pi \\
    \to x =  – 5\pi  + k12\pi \left( {k \in Z} \right)
   \end{array}\)

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