Question

Помогите решить, пожалуйста!!! Алгебра 10 класс

Answer 1

$\displaystyle\bf\\1)\\\\Cos\Big(\frac{x}{2} +\frac{\pi }{4} \Big)+1=0\\\\\\Cos\Big(\frac{x}{2} +\frac{\pi }{4} \Big)=-1\\\\\\\frac{x}{2} +\frac{\pi }{4} =\pi +2\pi n ,n\in Z\\\\\\\frac{x}{2} =\pi -\frac{\pi }{4} +2\pi n,n\in Z\\\\\\\frac{x}{2} =\frac{3\pi }{4} +2\pi n,n\in Z\\\\\\\boxed{x=\frac{3\pi }{2} +4\pi n,n\in Z}$

$\displaystyle\bf\\2)\\\\Sin^{2} x-2Cosx+2=0\\\\1-Cos^{2} x-2Cosx+2=0\\\\Cos^{2} x+2Cosx-3=0\\\\Cosx=m \ \ , \ \ -1\leq m\leq 1\\\\m^{2} +2m-3=0\\\\Teorema \ Vieta:\\\\m_{1} +m_{2} =-2\\\\m_{1} \cdot m_{2} =-3\\\\m_{1} =1\\\\m_{2} =-3 < -1-neyd\\\\Cosx=1\\\\\boxed{x=2\pi n ,n\in Z}$

Answer 2

Решить уравнения 1) cos(x/2 + π/4)+1=0; 2) sin^2 x - 2cosx+2=0.

Ответ:

$\Large \boldsymbol {} 1)\:x=\frac{3\pi }{2} +4\pi n, n\in Z\\\\2)\: x=2\pi n, n\in Z$

Формулы:

Если cos x = b и |b|≤1, то $\large \boldsymbol {}x=\arccos b +2\pi n, n\in Z$

$\Large \boldsymbol {} \sin^{2} \alpha +\cos^{2} \alpha =1$

Объяснение:

$\Large \boldsymbol{} \displastyle 1)\:\cos \left(\frac{x}{2} +\frac{\pi }{4} \right)+1=0\\\\\cos \left(\frac{x}{2} +\frac{\pi }{4} \right)=-1\\\\\frac{x}{2} +\frac{\pi }{4}=\arccos (-1) +2\pi n\\\\\frac{x}{2} =\pi +2\pi n-\frac{\pi }{4}\:\Big|_{}^{} *2\\\\x=2\pi +4\pi n-\frac{\pi }{2}\\\\x=\frac{4\pi }{2} +4\pi n-\frac{\pi }{2}\\\\x=\frac{3\pi }{2} +4\pi n, n\in Z$

$\Large \boldsymbol {} 2) \sin^{2} x-2\cos x+2=0$

С формулы $\large \boldsymbol {} \sin^{2} \alpha +\cos^{2} \alpha =1$ выразим cos^2 x и подставим вместо sin^2 x:

$\Large \boldsymbol {} 1-\cos^{2} x-2\cos x+2=0\\\\-\cos^{2} x-2\cos x+3=0\:\Big|_{}^{}*(-1)\\\\\cos^{2} x+2\cos x-3=0$

Введём замену cos^2 x = u

$\Large \boldsymbol {} u^{2} +2u-3=0\\u_1=\frac{-2+\sqrt{2^{2}-4*1*(-3) } }{2} =\frac{-2+4}{2} =1\\\\u_2=\frac{-2-\sqrt{2^{2}-4*1*(-3) } }{2} =\frac{-2-4}{2} =-3$

Возвращаемся к замене:

$\Large \boldsymbol {} \cos^{2} x=1 \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\cos^{2} x=-3\\\\\cos x=1 \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:x\notin R\\\\x=2\pi n, n\in Z$