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Ответ:

Maci189

26.12.2021 19:57

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

$ili\ \ \ \ \left[\begin{array}{l}\ \ x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+2\pi n\ ,\\x+\dfrac{\pi}{4}=\dfrac{3\pi }{4}+2\pi k\ ,\ k\in Z\end{array}\right\ \ \ Otvet:\ \ \left[\begin{array}{l}x=2\pi n\ ,\\x=\dfrac{\pi}{2}+2\pi k\ ,\ k\in Z\end{array}\right$

$2)\ \ \displaystyle sinx+cosx=1\ \Big|:\sqrt2\\\\\frac{1}{\sqrt2}\, sinx+\frac{1}{\sqrt2}\, cosx=\frac{1}{\sqrt2}\\\\\\sin\frac{\pi}{4}\cdot sinx+cos\frac{\pi}{4}\, cosx=\frac{\sqrt2}{2}\\\\\\cos\Big(x-\frac{\pi}{4}\Big)=\frac{\sqrt2}{2}\ \ \ ,\ \ \ cos\Big(x-\frac{\pi}{4}\Big)\ne cos\Big(2x-\frac{\pi }{2}\Big)\\\\\\x-\frac{\pi}{4}=\pm \frac{\pi }{4}+2\pi m\ ,\ m\in Z$

$Otvet:\ x=\dfrac{\pi}{4}\pm \dfrac{\pi }{4}+2\pi m\ ,\ m\in Z\ \ \ \Rightarrow \ \ \ \ x=\left[\begin{array}{l}\dfrac{\pi}{2}+2\pi m\ ,\ m\in Z\\2\pi l\ ,\ l\in Z\end{array}\right\ .$

$\displaystyle 3)\ \ cos\Big(2x-\frac{\pi}{2}\Big)=cos\Big(\frac{\pi}{2}-2x\Big)=sin2x=2\, sinx\cdot cosx\ne sinx+cosx\\\\\\cos\Big(2x-\frac{\pi}{2}\Big)=1\ \ \ \Rightarrow \ \ \ sin2x=1\ \ ,\ \ 2x=\frac{\pi}{2}+2\pi s\ ,\ s\in Z\ ,\\\\\\Otvet:\ \ x=\frac{\pi}{4}+\pi s\ ,\ s\in Z$