![\frac{1}{\sqrt{10} } * sin \frac{x}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%7B10%7D+%7D+%2A+sin+%5Cfrac%7Bx%7D%7B2%7D)
, cos x = -0,8, x ∈ (π; 3π/2)
![sin \frac{x}{2}](https://tex.z-dn.net/?f=sin+%5Cfrac%7Bx%7D%7B2%7D)
= (+/-)
![\sqrt{ \frac{1-cosx}{2} }](https://tex.z-dn.net/?f=+%5Csqrt%7B+%5Cfrac%7B1-cosx%7D%7B2%7D+%7D+)
x/2 ∈ (π/2; 3π/4) ⇒ угол x/2 лежит во II четверти ⇒
![sin \frac{x}{2}](https://tex.z-dn.net/?f=sin+%5Cfrac%7Bx%7D%7B2%7D)
> 0 ⇒
![sin \frac{x}{2}](https://tex.z-dn.net/?f=sin+%5Cfrac%7Bx%7D%7B2%7D)
=
![\sqrt{ \frac{1-cosx}{2} }](https://tex.z-dn.net/?f=%5Csqrt%7B+%5Cfrac%7B1-cosx%7D%7B2%7D+%7D+)
Окончательно:
![\frac{1}{\sqrt{10} } * sin \frac{x}{2}=\frac{1}{\sqrt{10}}*\sqrt{ \frac{1-cosx}{2} }=\sqrt{ \frac{1-cosx}{20}}=\sqrt{ \frac{1-(-0,8)}{20}}=\sqrt{ \frac{1,8}{20}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%7B10%7D+%7D+%2A+sin+%5Cfrac%7Bx%7D%7B2%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B10%7D%7D%2A%5Csqrt%7B+%5Cfrac%7B1-cosx%7D%7B2%7D+%7D%3D%5Csqrt%7B+%5Cfrac%7B1-cosx%7D%7B20%7D%7D%3D%5Csqrt%7B+%5Cfrac%7B1-%28-0%2C8%29%7D%7B20%7D%7D%3D%5Csqrt%7B+%5Cfrac%7B1%2C8%7D%7B20%7D%7D)
= 0,3
<span> (138+517)</span>²<span> >138</span>²<span>+517</span>², так как (138+517)²= 138²+517²+2·138·517
Cos(8x-π/8)≥√3*21/42
cos(8x-π/8)≥√3/2
2πn-π/6≤8х-π/8≤2πn+π/6, n∈Z
2πn-π/6+π/8≤8х≤2πn+π/6+π/8
2πn-π/24≤8х≤2πn+7π/24
πn/4-π/192<span>≤х≤πn/4+7π/192
</span>x∈[πn/4-π/192; <span>πn/4+7π/192</span>]
Если я правильно понял условие, 1-й замок имеет 5^6=15625 комбинаций, а второй 6^5=7776. Поэтому 1-й лучше..