Тригонометријске једначине. Једноставни примери.
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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 $ |
$tgx + \frac{1}{{\sqrt 3 }} = 0$ |
$tgx = \frac{1}{{\sqrt 3 }}$ |
|
$tgx = - \frac{1}{{\sqrt 3 }}$ |
$tgx = \frac{{\sqrt 3 }}{3}$ |
|
$tgx = - \frac{{\sqrt 3 }}{3}$ |
$x = \frac{\pi }{6} + k\pi ,k \in \mathbb{Z}$ |
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$x = \frac{{5\pi }}{6} + k\pi ,k \in \mathbb{Z}$ |