Задание 16.10

Question

$^{\circ}$ Решите уравнение:
$$1)\sin \left(\frac{\pi}{18}-8 x\right)=1$$
$$2) 2 \sin \left(\frac{x}{5}-4\right)+1=0$$

Gimli · Accepted Answer

\begin{enumerate}
\item 
$$ \sin \left(\frac{\pi}{18}-8 x\right)=1 $$
$$\frac{\pi}{18}-8 x=(-1)^{n} \cdot \arcsin 1+\pi n$$
$$\frac{\pi}{18}-8 x=(-1)^{n} \cdot \frac{\pi}{2}+\pi n $$
$$\cdot 8 x=\frac{\pi}{18}-(-1)^{n} \cdot \frac{\pi}{2}-\pi n$$
$$x=\frac{\pi}{144}+(-1)^{n+1} \cdot \frac{\pi}{16}-\frac{\pi n}{8}, n \in \mathbb{Z} $$

\item
$$2 \sin \left(\frac{x}{5}-4\right)+1=0 \quad \sin \left(\frac{x}{5}-4\right)=-\frac{1}{2} $$
$$\frac{x}{5}-4=(-1)^{n} \arcsin \left(-\frac{1}{2}\right)+\pi n $$
$$\frac{x}{5}-4=(-1)^{n+1} \arcsin \frac{1}{2}+14 n$$
$$\frac{x}{5}=(-1)^{n+1} \arcsin \frac{1}{2}+\pi n+4 $$
$$x=5\left((-1)^{n+1} \arcsin \frac{1}{2}+\pi n+4\right) $$
$$x=5\left((-1)^{n+1} \cdot \frac{\pi}{6}+\pi n+4\right)$$
$$x=5(-1)^{n+1} \cdot \frac{5 \pi}{6}+5 \pi n+20 ,n \in \mathbb{Z}$$

\end{enumerate}

Задание 16.10

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Задание 16.10

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