Question

уравнения. 90 баллов....

Answer 1

Ответ:

$\displaystyle 7)\ \ cosx+sinx=1\ \Big|:\sqrt2\\\\\frac{1}{\sqrt2}\cdot cosx+\frac{1}{\sqrt2}\cdot sinx=\frac{1}{\sqrt2}\\\\cos\frac{\pi}{4}\cdot cosx+sin\frac{\pi}{4}\cdot sinx=\frac{\sqrt2}{2}\\\\cos\Big(x-\dfrac{\pi }{4}\Big)=\frac{\sqrt2}{2}\\\\x-\dfrac{\pi }{4}=\pm \frac{\pi}{4}+2\pi n\ ,\ n\in Z\\\\x=\dfrac{\pi }{4}\pm \frac{\pi}{4}+2\pi n\ ,\ n\in Z$

${}\ ili\ \ \ \ \left[\begin{array}{l}x=2\pi n\ ,\ n\in Z\\x=\dfrac{\pi}{2}+2\pi k\ ,\ k\in Z\end{array}\right\\\\\\Otvet:\ \ x_1=2\pi n\ ,\ \ x_2=\dfrac{\pi }{2}+2\pi k\ \ ,\ n,k\in Z$

$8)\ \ \sqrt3sinx=sin2x\\\\\sqrt3sinx-2sinx\cdot cosx=0\\\\sinx\cdot (\sqrt3-2\, cosx)=0\\\\a)\ \ sinx=0\ \ ,\ \ x=\pi n\ ,\ n\in Z\\\\b)\ \ cosx=\dfrac{\sqrt3}{2}\ \ ,\ \ x=\pm \dfrac{\pi}{6}+2\pi k\ ,\ k\in Z\\\\Otvet:\ \ x_1=\pi n\ \ ,\ \ x_2=\pm \dfrac{\pi }{6}+2\pi k\ \ ,\ n,k\in Z\ .$

$9)\ \ cos10x=cos8x\\\\cos10x-cos8x=0\\\\-2sinx\cdot sin6x=0\\\\a)\ \ sinx=0\ ,\ \ x=\pi n\ ,\ n\in Z\\\\b)\ \ cos6x=0\ \ ,\ \ 6x=\dfrac{\pi}{2}+\pi k\ \ ,\ \ x=\dfrac{\pi}{12}+\dfrac{\pi k}{6}\ ,\ k\in Z\\\\Otvet:\ x_1=\pi n\ \ ,\ x_2=\dfrac{\pi}{12}+\dfrac{\pi k}{6}\ ,\ n,k\in Z\ .$