Задание 15.3

Question

15.3. $^{\circ}$ Решите уравнение:
1) $\cos 3 x=-\frac{1}{2}$
3) $\cos 6 x=1$
5) $\cos 9 x=-\frac{1}{5}$
2) $\cos \frac{5}{6} x=\frac{\sqrt{3}}{2}$
4) $\cos \frac{2 \pi x}{3}=0$;
6) $\cos \left(-\frac{x}{3}\right)=\frac{\sqrt{3}}{3}$

Gimli · Accepted Answer

1) $$
\begin{array}{l}
\cos 3x =-\dfrac{1}{2}\[10pt]
3 x=\pm \arccos \left(-\dfrac{1}{2}\right)+2 \pi n \[10pt]
3 x=\pm \dfrac{2}{3} \pi+2 \pi n \[10pt]
x=\pm \dfrac{2}{9} \pi+\dfrac{2}{3} \pi n, \quad n \in \mathbb{Z}
\end{array}
$$

2) 
$$
\begin{array}{l}
\cos \dfrac{5}{6} x=\dfrac{\sqrt{3}}{2} \[10pt]
\dfrac{5}{6} x=\pm \arccos \dfrac{\sqrt{3}}{2}+2 \pi n \[10pt]
\dfrac{5}{6} x=\pm \dfrac{\pi}{6}+2 \pi n \[10pt]
x=\pm \dfrac{\pi}{5}+\dfrac{12}{5} \pi n, \quad n \in \mathbb{Z}
\end{array}
$$

3)
$$
\begin{array}{l}
\cos 6 x=1 \[10pt]
6 x=\pm \arccos 1+2 \pi n \[10pt]
6 x=2 \pi n \[10pt]
x=\dfrac{1}{3} \pi n , \quad n \in \mathbb{Z}
\end{array}
$$

4)
$$
\begin{array}{l}
\cos \dfrac{2 \pi x}{3}=0 \[10pt]
\dfrac{2 \pi x}{3}=\pm \arccos 0+2 \pi n \[10pt]
\dfrac{2 \pi x}{3}=\pm \dfrac{\pi}{2}+2 \pi n \[10pt]
x=\pm \dfrac{3}{4}+3 n , \quad n \in \mathbb{Z}
\end{array}
$$

5)
$$
\begin{array}{l}
\cos 9 x=-\dfrac{1}{5} \[10pt]
9 x=\pm \arccos \left(-\dfrac{1}{5}\right)+2 \pi n \[10pt]
x=\dfrac{\pm\arccos \left(-\dfrac{1}{5}\right)}{9}+\dfrac{2}{9} \pi n, \quad n \in \mathbb{Z}
\end{array}
$$

6) 
$$
\begin{array}{l}
\cos \left(-\dfrac{x}{3}\right)=\dfrac{\sqrt{3}}{3} \[10pt]
-\dfrac{x}{3}=\pm \arccos \dfrac{\sqrt{3}}{3}+2 \pi n \[10pt]
x=\pm 3 \cdot \arccos \dfrac{\sqrt{3}}{3}+6 \pi n, \quad n \in \mathbb{Z}
\end{array}
$$

Задание 15.3

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Задание 15.3

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