Задание 15.5

Question

15.5. $^{\circ}$ Решите уравнение:
1) $\cos \left(x+\frac{\pi}{6}\right)=\frac{\sqrt{2}}{2}$
3) $\cos \left(\frac{x}{6}-2\right)=-1$
2) $\cos \left(\frac{\pi}{4}-x\right)=-\frac{\sqrt{3}}{2}$
4) $2 \cos \left(\frac{\pi}{8}-3 x\right)+1=0$

Gimli · Accepted Answer

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

2)
$$
\begin{array}{l}
	ext { 2) }\cos \left(\frac{\pi}{4}-xight)=-\frac{\sqrt{3}}{2} \
\frac{\pi}{4}-x=\pm \arccos \left(-\frac{\sqrt{3}}{2}ight)+2 \pi n \
\frac{\pi}{4}-x=\pm \frac{5 \pi}{6}+2 \pi n \
x=\pm \frac{5 \pi}{6}+\frac{\pi}{4}+2 \pi n, \quad n \in \mathbb{Z}\end{array}
$$

3)
$$
\begin{array}{l}
\cos \left(\frac{x}{6}-2ight)=-1 \
\frac{x}{6}-2=\pm \arccos (-1)+2 \pi n \
\frac{x}{6}=\pm \pi+2+2 \pi n \
x=\pm 6 \pi+12 \pi n+12, \quad n \in \mathbb{Z}
\end{array}
$$

4)
$$
\begin{array}{l}
\cos \left(\frac{\pi}{8}-3 xight)+1=0 \
2 \cos \left(\frac{\pi}{8}-3 xight)=-1 \
\cos \left(\frac{\pi}{8}-3 xight)=-\frac{1}{2} \
\frac{\pi}{8}-3 x=\pm \arccos \left(-\frac{1}{2}ight)+2 \pi n \
-3 x=\pm \frac{2}{3} \pi-\frac{\pi}{8}+2 \pi n \
x=\pm \frac{2}{9} \pi+\frac{\pi}{24}+\frac{2}{3} \pi n, \quad n \in \mathbb{Z}
\end{array}
$$

Задание 15.5

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Задание 15.5

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