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  • Voxman
  • главный мозг
2013-06-09T14:41:23+00:00

 

|\sqrt{\pi^2 - x^2} - 10| + \sqrt{2}\sin(x + \frac{\pi}{4}) + \sqrt{\pi^2 - x^2} = 11 - sinx*cosx\\\\ \pi^2 - x^2 \geq 0\\\\ -x^2 \geq -\pi^2\\\\ x^2 \leq \pi^2\\\\ x \in [-\pi;\pi]\\\\ \downarrow\\\\ \sqrt{\pi^2 - x^2} - 10 \leq 0

 

 

-\underline{\sqrt{\pi^2 - x^2}} + 10 + \sqrt{2}\sin(x + \frac{\pi}{4}) + \underline{\sqrt{\pi^2 - x^2}} = 11 - sinx*cosx\\\\ 10 + \sqrt{2}\sin(x + \frac{\pi}{4}) = 11 - sinx*cosx\\\\ \sqrt{2}\sin(x + \frac{\pi}{4}) + sinx*cosx = 1\\\\ \boxed{ \sqrt{2}\sin(x + \frac{\pi}{4}) = sinx + cosx }\\\\\\ sinx + cosx + sinx*cosx - 1 = 0\\\\ sinx + cosx = y\\\\ y^2 = sin^2x+2sinxcosx+cos^2x = 2sinxcosx + 1\\\\ sinxcosx = \frac{y^2 - 1}{2}\\\\

 

y + \frac{y^2 - 1}{2} - 1 = 0 | * 2\\\\ y^2 - 1 + 2y - 2 = 0\\\\ y^2 + 2y - 3 = 0\\\\ y_1*y_2 = -3 = -3*1\\\\ y_1+y_2 = -2 = -3 + 1\\\\ y_1 = -3, \ \underline{y_2 = 1}\\\\ cosx + sinx = 1\\\\ \sqrt{2}sin(x + \frac{\pi}{4}) = 1\\\\ sin(x + \frac{\pi}{4}) = \frac{\sqrt{2}}{2}\\\\ x + \frac{\pi}{4} = \frac{\pi}{4} + 2 \pi n, \ n \in Z\\\\ x + \frac{\pi}{4} = \frac{3\pi}{4} + 2\pi n, \ n \in Z\\\\ x = 2 \pi n, \ n \in Z\\\\ x = \frac{\pi}{2} + 2\pi n, \ n \in Z\\\\

 

В область допустимых значений попадают корни 0 и \frac{\pi}{2}, которые равны, соответственно: 0^{\circ}, 90^{\circ}. Их среднее арифметическое равно:

 

\boxed{ \frac{0^{\circ} + 90^{\circ}}{2} = 45^{\circ} }