Giải pt lượng giác sau a . tan^2 2x . tan^2 3x = 1 b. cot (3x + 2π/3 ). tan (x – π/3 ) = 1 Giúp giùm e vs

Question

Giải pt lượng giác sau
a . tan^2 2x . tan^2 3x = 1
b. cot (3x + 2π/3 ). tan (x – π/3 ) = 1
Giúp giùm e vs

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RuslanHeatt 4 years 2020-10-16T14:34:05+00:00 2 Answers 317 views 0

Answers ( )

  1. Orla
    0
    2020-10-16T14:36:04+00:00
    Reply

    \[\begin{array}{l}
    a)\,\,{\tan ^2}2x.{\tan ^2}3x = 1\,\,\,\left( * \right)\\
    DK:\,\,\left\{ \begin{array}{l}
    \cos 2x \ne 0\\
    \cos 3x \ne 0
    \end{array} \right..\\
    \Rightarrow \left( * \right) \Leftrightarrow \frac{{{{\sin }^2}2x}}{{{{\cos }^2}2x}}.\frac{{{{\sin }^2}3x}}{{{{\cos }^2}3x}} = 1\\
    \Leftrightarrow {\left( {\sin 2x.\sin 3x} \right)^2} = {\left( {\cos 2x.\cos 3x} \right)^2}\\
    \Leftrightarrow {\left[ {\frac{1}{2}\left( {\cos x – \cos 5x} \right)} \right]^2} = {\left[ {\frac{1}{2}\left( {\cos 5x + \cos x} \right)} \right]^2}\\
    \Leftrightarrow {\left( {\cos x – \cos 5x} \right)^2} = {\left( {\cos x + \cos 5x} \right)^2}\\
    \Leftrightarrow – 2\cos x.\cos 5x = 2\cos x.\cos 5x\\
    \Leftrightarrow \cos x.\cos 5x = 0\\
    \Leftrightarrow \left[ \begin{array}{l}
    \cos x = 0\\
    \cos 5x = 0
    \end{array} \right..\\
    b)\,\,\,\cot \left( {3x + \frac{{2\pi }}{3}} \right).tan\left( {x – \frac{\pi }{3}} \right) = 1\,\,\,\left( * \right)\\
    DK:\,\,\,\left\{ \begin{array}{l}
    \cos \left( {x – \frac{\pi }{3}} \right) \ne 0\\
    \sin \left( {3x\_\frac{{2\pi }}{3}} \right) \ne 0
    \end{array} \right.\\
    \left( * \right) \Leftrightarrow \frac{{\cos \left( {3x + \frac{{2\pi }}{3}} \right)}}{{\sin \left( {3x\_\frac{{2\pi }}{3}} \right)}}.\frac{{\sin \left( {x – \frac{\pi }{3}} \right)}}{{\cos \left( {x – \frac{\pi }{3}} \right)}} = 1\\
    \Leftrightarrow \cos \left( {3x + \frac{{2\pi }}{3}} \right).\sin \left( {x – \frac{\pi }{3}} \right) = \sin \left( {3x – \frac{{2\pi }}{3}} \right).\cos \left( {x – \frac{\pi }{3}} \right)\\
    \Leftrightarrow \sin \left( {x – \frac{\pi }{3} – 3x – \frac{{2\pi }}{3}} \right) + \sin \left( {x – \frac{\pi }{3} + 3x + \frac{{2\pi }}{3}} \right) = \sin \left( {3x – \frac{{2\pi }}{3} – x + \frac{\pi }{3}} \right) + \sin \left( {3x – \frac{{2\pi }}{3} + x – \frac{\pi }{3}} \right)\\
    \Leftrightarrow \sin \left( { – 2x – \pi } \right) + \sin \left( {4x + \frac{\pi }{3}} \right) = \sin \left( {2x – \frac{\pi }{3}} \right) + \sin \left( {4x – \pi } \right)\\
    \Leftrightarrow \sin \left( {2x + \pi } \right) + \sin \left( {4x + \frac{\pi }{3}} \right) = \sin \left( {2x – \frac{\pi }{3}} \right) – \sin \left( {\pi – 4x} \right)\\
    \Leftrightarrow – \sin 2x + \sin 4x.\cos \frac{\pi }{3} + \cos 4x.\sin \frac{\pi }{3} = \sin 2x.cos\frac{\pi }{3} – \cos 2x.\sin \frac{\pi }{3} – \sin 4x\\
    \Leftrightarrow \frac{1}{2}\sin 4x + \frac{{\sqrt 3 }}{2}\cos 4x – \sin 2x = \frac{1}{2}\sin 2x – \frac{{\sqrt 3 }}{2}\cos 2x – \sin 4x\\
    \Leftrightarrow \frac{3}{2}\sin 4x + \frac{{\sqrt 3 }}{2}\cos 4x = \frac{3}{2}\sin 2x – \frac{{\sqrt 3 }}{2}\cos 2x\\
    \Leftrightarrow \frac{{\sqrt 3 }}{2}\sin 4x + \frac{1}{2}\cos 4x = \frac{{\sqrt 3 }}{2}\sin 2x – \frac{1}{2}\cos 2x\\
    \Leftrightarrow \sin \left( {4x + \frac{\pi }{6}} \right) = \sin \left( {2x – \frac{\pi }{6}} \right).
    \end{array}\]
    Em giải các phương trình lượng giác cơ bản nhé.

  2. nguyetanh
    0
    2020-10-16T14:36:14+00:00
    Reply

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