Задание 15.6

Question

15.6. $^{\circ}$ Решите уравнение:
1) $\cos \left(\frac{\pi}{9}-4 x\right)=1;$
2) $\sqrt{2} \cos \left(\frac{x}{2}+3\right)+1=0$

Gimli · Accepted Answer

\begin{enumerate}
\item$$
\begin{array}{l}
 \cos \left(\frac{\pi}{9}-4 xight)=1 \
\frac{\pi}{9}-4 x=\pm\arccos 1 +2\pi n\
4 x=\frac{\pi}{9}\pm\arccos 1 +2\pi n\
4 x=\frac{\pi}{9} +2\pi n\
x=\frac{\pi}{36}+\frac{1}{2} \pi n, \quad n \in \mathbb{Z}
\end{array}
$$
ewline
\item$$
\begin{array}{l}
\sqrt{2} \cos \left(\frac{x}{2}+3ight)+1=0 \
\cos \left(\frac{x}{2}+3ight)=\frac{-1}{\sqrt{2}} \
\frac{x}{2}+3=\pm\arccos \left(-\frac{1}{\sqrt{2}}ight)+2 \pi  n \
\frac{x}{2}+3=\pm\frac{3 \pi}{4}+2 \pi  n \
\frac{x}{2}=\pm\frac{3 \pi}{4}-3+2 \pi  n\
x=\pm\frac{3 \pi}{2}-6 + 4 \pi n, \quad n \in \mathbb{Z}
\end{array}
$$
\end{enumerate}

Задание 15.6

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Задание 15.6

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