Задание 16.14

Question

16.14. Найдите наибольший отрицательный корень уравнения $\sin \left(3 x-\frac{\pi}{15}\right)=-1$

Gimli · Accepted Answer

$$\sin \left(3 x-\frac{\pi}{15}ight)=-1$$
$$3 x-\frac{\pi}{15}=(-1)^{k} \arcsin (-1)+\pi k$$
$$3 x-\frac{\pi}{15}=(-1)^{k+1} \cdot \frac{\pi}{2}+\pi k $$
$$3 x=(-1)^{k+1} \cdot \frac{\pi}{2}+\pi k+\frac{\pi}{15} $$
$$x=(-1)^{k+1} \cdot \frac{\pi}{6}+\frac{\pi k}{3}+\frac{\pi}{45}$$

Разберем оба случая:
\begin{enumerate}
\item  $k$ - нечетное
$$

x=\frac{\pi}{6}+\frac{\pi k}{3}+\frac{\pi}{45}=\frac{17}{90} \pi+\frac{\pi k}{3} $$
$$\frac{17}{90} \pi+\frac{\pi k}{3}<0 $$
$$\frac{\pi k}{3}<-\frac{17}{90} \pi $$
$$k<-\frac{17}{30}, k \in \mathbb{Z} $$
$$k=-1$$ 
$$x=\frac{17}{90} \pi-\frac{\pi}{3}=-\frac{13 \pi}{90}$$

\item $k$ - четное
$$x=-\frac{\pi}{6}+\frac{\pi }{45}+\frac{\pi K}{3}=-\frac{13}{90} \pi+\frac{\pi K}{3} $$
$$-\frac{13}{90} \pi+\frac{\pi K}{3}<0 $$
$$\frac{\pi K}{3}<\frac{13}{90} $$
$$k<\frac{39}{90}, k \in \mathbb{Z} $$
$$k=0 $$
$$x=-\frac{13}{90} \pi $$
\end{enumerate}
	extbf{Ответ}:$-\frac{13}{90} \pi$

Задание 16.14

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Задание 16.14

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