Монгол Бодлогын Сан

Заавар: $\sqrt{2}\cos 8x+\sqrt{2}\sin 8x=-1\Rightarrow \dfrac{\sqrt{2}}2\cos 8x+\dfrac{\sqrt{2}}2\sin 8x=-\dfrac12$ буюу $\sin\left(8x+\dfrac{\pi}4\right)=-\dfrac12$ бод.

Бодолт: $$\sqrt{2}\cos 8x+\sqrt{2}\sin 8x=-1\Rightarrow \dfrac{\sqrt{2}}2\cos 8x+\dfrac{\sqrt{2}}2\sin 8x=-\dfrac12$$ буюу $$\sin\left(8x+\dfrac{\pi}4\right)=-\dfrac12\Rightarrow 8x+\dfrac{\pi}4=(-1)^{k+1}\dfrac{\pi}{6}+\pi k\Rightarrow x=\dfrac{12 k-3+2\cdot(-1)^{k+1}}{96}\pi$$ болно. $\dfrac{3\pi}{8}< x< \dfrac{7\pi}{10}$ нөхцлийг хангах шийдүүдийг олъё. $$\dfrac38<\dfrac{12 k-3+2\cdot(-1)^{k+1}}{96}<\dfrac7{10}\Rightarrow k=4, 5, 6$$ болно. $k=4$ үед $x_4=\dfrac{43}{96}\pi$, $k=6$ үед $x_6=\dfrac{67}{96}\pi\Rightarrow \dfrac{x_4}{x_6}=\dfrac{43}{67}.$