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Тригонометријске једначине 2

Задаци

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Решити тригонометријску једначину.

Пр.6)   $\sin \left( {2x + \frac{\pi }{3}} \right) = \sin \left( {2x + \frac{\pi }{4}} \right)$

Пр.7)   $\sin x + \cos x - 1 + \sin x\cos x$

Пр.8)   $t{g^2}x = \frac{1}{3}$

Пр.6)   $\sin \left( {2x + \frac{\pi }{3}} \right) = \sin \left( {2x + \frac{\pi }{4}} \right)$

\[\begin{gathered}
  \sin \left( {2x + \frac{\pi }{3}} \right) = \sin \left( {2x + \frac{\pi }{4}} \right) \hfill \\
  \sin \left( {2x + \frac{\pi }{3}} \right) - \sin \left( {2x + \frac{\pi }{4}} \right) = 0 \hfill \\
  2\cos \frac{{\left( {2x + \frac{\pi }{3}} \right) + \left( {2x + \frac{\pi }{4}} \right)}}{2}\sin \frac{{\left( {2x + \frac{\pi }{3}} \right) - \left( {2x + \frac{\pi }{4}} \right)}}{2} = 0 \hfill \\
  2\cos \frac{{4x + \frac{{7\pi }}{{12}}}}{2}\sin \frac{{\frac{\pi }{{12}}}}{2} = 0 \hfill \\
  2\cos \left( {2x + \frac{{7\pi }}{{24}}} \right)\sin \frac{\pi }{{24}} = 0 \hfill \\
  \cos \left( {2x + \frac{{7\pi }}{{24}}} \right) = 0 \hfill \\
  2x + \frac{{7\pi }}{{24}} = \frac{\pi }{2} + k\pi ,k \in \mathbb{Z} \hfill \\
  2x = \frac{\pi }{2} - \frac{{7\pi }}{{24}} + k\pi ,k \in \mathbb{Z} \hfill \\
  2x = \frac{{5\pi }}{{24}} + k\pi ,k \in \mathbb{Z}\left| {:2} \right. \hfill \\
  x = \frac{{5\pi }}{{48}} + \frac{{\pi k}}{2},k \in \mathbb{Z} \hfill \\
\end{gathered} \]

Пр.7)   $\sin x + \cos x = 1 + \sin x\cos x$

\[\begin{gathered}
  \sin x + \cos x = 1 + \sin x\cos x \hfill \\
  \sin x + \cos x - 1 - \sin x\cos x = 0 \hfill \\
  \sin x\left( {1 - \cos x} \right) - \left( {1 - \cos x} \right) = 0 \hfill \\
  \left( {1 - \cos x} \right)\left( {\sin x - 1} \right) = 0 \hfill \\
  1 - \cos x = 0 \hfill \\
  \cos x = 1 \hfill \\
  x = 2k\pi ,k \in \mathbb{Z} \hfill \\
  \sin x - 1 = 0 \hfill \\
  \sin x = 1 \hfill \\
  x = \frac{\pi }{2} + 2k\pi ,k \in \mathbb{Z} \hfill \\
\end{gathered} \]

Пр.8)  

$t{g^2}x = \frac{1}{3}$

$t{g^2}x - \frac{1}{3} = 0$

$t{g^2}x - {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} = 0$

$\left( {tgx - \frac{1}{{\sqrt 3 }}} \right)\left( {tgx + \frac{1}{{\sqrt 3 }}} \right) = 0$

$tgx - \frac{1}{{\sqrt 3 }} = 0$$ \cup $   $tgx + \frac{1}{{\sqrt 3 }} = 0$
$tgx = \frac{1}{{\sqrt 3 }}$    $tgx =  - \frac{1}{{\sqrt 3 }}$
$tgx = \frac{{\sqrt 3 }}{3}$    $tgx =  - \frac{{\sqrt 3 }}{3}$
$x = \frac{\pi }{6} + k\pi ,k \in \mathbb{Z}$    $x = \frac{{5\pi }}{6} + k\pi ,k \in \mathbb{Z}$
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