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

Question

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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Cherry 8 tháng 2020-10-31T22:15:42+00:00 1 Answers 57 views 0

Answers ( )

  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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