Предмет: Алгебра, автор: Аноним

задание номер 133 алгебра

Приложения:

Ответы

Автор ответа: NNNLLL54

$1)\; \; \sqrt3cosx+sinx=\sqrt2\; |:2\qquad \Big (2=\sqrt{(\sqrt3)^2+1^2}\Big )\\\\\frac{\sqrt3}{2}cosx+\frac{1}{2}sinx=\frac{\sqrt2}{2}\\\\sin\frac{\pi}{3}\cdot cosx+cos\frac{\pi}{3}sinx=\frac{\sqrt3}{2}\\\\sin(x+\frac{\pi}{3})=\frac{\sqrt3}{2}\\\\x+\frac{\pi}{3}=(-1)^{n}\cdot \frac{\pi}{3}+\pi n\; ,\; n\in Z\\\\x=(-1)^{n}\cdot \frac{\pi}{3}-\frac{\pi}{3}+\pi n\; ,\; n\in Z$

$2)\; \; \sqrt2cosx-\sqrt2sinx=1\; |:2\qquad \Big (2=\sqrt{(\sqrt2)^2+(\sqrt2)^2}\Big )\\\\\frac{\sqrt2}{2}cosx-\frac{\sqrt2}{2}sinx=\frac{1}{2}\\\\sin\frac{\pi}{4}\cdot cosx-cos\frac{\pi}{4}\cdot sinx=\frac{1}{2}\\\\sin(\frac{\pi}{4}-x)=\frac{1}{2}\\\\sin(x-\frac{\pi}{4})=-\frac{1}{2}\\\\x-\frac{\pi}{4}=(-1)^{n}\cdot (-\frac{\pi}{6})+\pi n\; ,\; n\in Z\\\\x=(-1)^{n+1}\cdot \frac{\pi}{6}+\frac{\pi }{4}+\pi n\; ,\; n\in Z$

$3)\; \; sinx-cosx=-\sqrt2\; |:\sqrt2\qquad \Big (\sqrt2=\sqrt{1^2+1^2}\Big )\\\\\frac{1}{\sqrt2}sinx-\frac{1}{\sqrt2}cosx=-1\\\\\frac{\sqrt2}{2}sinx-\frac{\sqrt2}{2}cosx=-1\\\\cos\frac{\pi}{4}\cdot sinx-sin\frac{\pi}{4}\cdot cosx=-1\\\\sin(x-\frac{\pi}{4})=-1\\\\x-\frac{\pi}{4}=-\frac{\pi}{2}+2\pi n\; ,\; n\in Z\\\\x=-\frac{\pi}{4}+2\pi n\; ,\; n\in Z$