Задание 16.9

Question

16.9. ${ }^{\circ}$ Решите уравнение:
$$1) \sin \left(x-\frac{\pi}{6}\right)=\frac{\sqrt{2}}{2}$$
$$3) \sin \left(\frac{x}{3}+1\right)=-1$$
$$2) \sin \left(\frac{\pi}{8}-x\right)=\frac{\sqrt{3}}{2}$$
$$4) \sqrt{2} \sin \left(\frac{\pi}{12}-3 x\right)-1=0$$

Gimli · Accepted Answer

1) $$\sin \left(x-\frac{\pi}{6}ight)=\frac{\sqrt{2}}{2}$$
$$x-\frac{\pi}{6}=(-1)^{n} \operatorname{arcsin} \frac{\sqrt{2}}{2}+\pi n$$
$$x-\frac{\pi}{6}=(-1)^{n} \cdot \frac{\pi}{4}+\pi n$$
$$x=(-1)^{n} \frac{\pi}{4}+\frac{\pi}{6}+\pi n$$
2) $$\sin \left(\frac{\pi}{8}-xight)=\frac{\sqrt{3}}{2}$$
$$-\sin \left(x-\frac{\pi}{8}ight)=\frac{\sqrt{3}}{2}$$
$$\sin \left(x-\frac{\pi}{8}ight)=-\frac{\sqrt{3}}{2}$$
$$x-\frac{\pi}{8}=(-1)^{n} \cdot \arcsin \left(-\frac{\sqrt{3}}{2}ight)+\pi n$$
$$x-\frac{\pi}{8}=(-1)^{n+1} \cdot \frac{\pi}{3}+\pi n$$
$$x=(-1)^{n+1} \frac{\pi}{3}+\frac{\pi}{8}+\pi n, n \in Z$$
3) $$\sin \left(\frac{x}{3}+1ight)=-1$$
 $$\frac{x}{3}+1=(-1)^{n} \arcsin (-1)+\pi n$$ 
$$\frac{x}{3}+1=(-1)^{n+1} \frac{\pi}{2}+\pi n$$
 $$\frac{x}{3}=(-1)^{n+1} \frac{\pi}{2}+\pi n-1$$ 
$$x=(-1)^{n+1} \cdot \frac{3 \pi}{2}+3 \pi n-3, n \in Z$$
$$

\begin{array}{l}
	ext { 4) } \sqrt{2} \sin \left(\frac{\pi}{12}-3 xight)-1=0 \
\sqrt{2} \sin \left(\frac{\pi}{12}-3 xight)=1 \
\sin \left(\frac{\pi}{12}-3 xight)=\frac{1}{\sqrt{2}} \
\frac{\pi}{12}-3 x=(-1)^{n} \arcsin \frac{1}{\sqrt{2}}+\pi n \
-3 x=(-1)^{n} \arcsin \frac{\sqrt{2}}{2}-\frac{\pi}{12}+\pi n \
-3 x=(-1)^{n} \frac{\pi}{4}-\frac{\pi}{12}+\pi n \
x=(-1)^{n+1} \frac{\pi}{12}+\frac{\pi}{36}+\frac{\pi n}{3}, n \in 2
\end{array}
$$

Задание 16.9

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Задание 16.9

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