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Тригонометријске једначине 1

Задаци

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Решити тригонометријску једначину.

Пр.1)   $2\cos x - \sqrt 3  = 0$

Пр.2)   $\sin 2x + 1 = 0$

Пр.3)   $2\cos \frac{x}{8} = \sqrt 2 $

Пр.4)   $3tg\frac{x}{2} + \sqrt 3  = 0$

Пр.5)   $2ctg\left( {x - \frac{\pi }{2}} \right) + 2 = 0$

Пр.1)   $2\cos x - \sqrt 3  = 0$

\[\begin{gathered}
  2\cos x = \sqrt 3  \hfill \\
  \cos x = \frac{{\sqrt 3 }}{2} \hfill \\
  x = \frac{\pi }{6} + 2\pi k, \hfill \\
  x = \frac{{11\pi }}{6} + 2\pi k,k \in \mathbb{Z} \hfill \\
\end{gathered} \]

Пр.2)   $\sin 2x + 1 = 0$

\[\begin{gathered}
  \sin 2x =  - 1 \hfill \\
  2x = \frac{{3\pi }}{6} + 2\pi k,k \in \mathbb{Z}\left| {:2} \right. \hfill \\
  x = \frac{{3\pi }}{4} + \pi k,k \in \mathbb{Z} \hfill \\
\end{gathered} \]

Пр.3)   $2\cos \frac{x}{8} = \sqrt 2 $

\[\begin{gathered}
  2\cos \frac{x}{8} = \sqrt 2 \left| {:2} \right. \hfill \\
  \cos \frac{x}{8} = \frac{{\sqrt 2 }}{2} \hfill \\
\end{gathered} \]

$\frac{x}{8} = \frac{\pi }{4} + 2\pi k,k \in \mathbb{Z}\left| { \cdot 8} \right.$   $\frac{x}{8} = \frac{{7\pi }}{4} + 2\pi k,k \in \mathbb{Z}\left| { \cdot 8} \right.$
$x = 2\pi  + 16k\pi,k \in \mathbb{Z} $   $x = 14\pi  + 16k\pi ,k \in \mathbb{Z}$

 

Пр.4)   $3tg\frac{x}{2} + \sqrt 3  = 0$

\[\begin{gathered}
  3tg\frac{x}{2} =  - \sqrt 3  \hfill \\
  tg\frac{x}{2} = \frac{{ - \sqrt 3 }}{3} \hfill \\
  \frac{x}{2} = \frac{{5\pi }}{6} + k\pi ,k \in \mathbb{Z}\left| { \cdot 2} \right. \hfill \\
  x = \frac{{5\pi }}{3} + 2k\pi ,k \in \mathbb{Z} \hfill \\
\end{gathered} \]

Пр.5)   $2ctg\left( {x - \frac{\pi }{2}} \right) + 2 = 0$

\[\begin{gathered}
  2ctg\left( {x - \frac{\pi }{2}} \right) =  - 2\left| {:2} \right. \hfill \\
  ctg\left( {x - \frac{\pi }{2}} \right) =  - 1 \hfill \\
  x - \frac{\pi }{2} = \frac{{3\pi }}{4} + k\pi ,k \in \mathbb{Z} \hfill \\
  x = \frac{{5\pi }}{4} + k\pi ,k \in \mathbb{Z} \hfill \\
\end{gathered} \]

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