Question

уравнения. 90 баллов. ..

Answer 1

Ответ:

$10)\ \ cos3x+sin6x=0\\\\cos3x+2sin3x\cdot cos3x=0\\\\cos3x\cdot (1+2sin3x)=0\\\\a)\ \ cos3x=0\ \ ,\ \ 3x=\dfrac{\pi }{2}+\pi n\ \ ,\ \ x=\dfrac{\pi}{6}+\dfrac{\pi n}{3}\ ,\ \ n\in Z\\\\b)\ \ sin3x=-\dfrac{1}{2}\ \ ,\ \ 3x=(-1)^{n}\cdot \Big(-\dfrac{\pi}{6}\Big)+\pi k\ \ ,\ \ 3x=(-1)^{n+1}\cdot \dfrac{\pi}{6}+\pi k\ ,\\\\x=(-1)^{n+1}\cdot \dfrac{\pi}{18}+\dfrac{\pi k}{3}\ ,\ k\in Z\\\\Otvet:\ x=\dfrac{\pi}{6}+\dfrac{\pi n}{3}\ ,\ \ x=(-1)^{n+1}\cdot \dfrac{\pi}{18}+\dfrac{\pi k}{3}\ ,\ n,k\in Z\ .$

$11)\ \ cosx+sinx=0\ \Big|:cosx\ne 0\\\\1+tgx=0\ \ ,\ \ tgx=-1\ \ ,\\\\x=-\dfrac{\pi}{4}+\pi k\ ,\ k\in Z\ \ -\ \ otvet$

$\displaystyle 12)\ \ cos3x+sin3x=1\ \Big|:\sqrt2\\\\\frac{1}{\sqrt2}\cdot cos3x+\frac{1}{\sqrt2}\cdot sin3x=\frac{1}{\sqrt2}\\\\cos\frac{\pi}{4}\cdot cos3x+sin\frac{\pi}{4}\cdot sin3x=\frac{\sqrt2}{2}\\\\cos\Big(3x-\dfrac{\pi }{4}\Big)=\frac{\sqrt2}{2}\\\\3x-\dfrac{\pi }{4}=\pm \frac{\pi}{4}+2\pi n\ ,\ n\in Z\\\\3x=\dfrac{\pi }{4}\pm \frac{\pi}{4}+2\pi n\ ,\ n\in Z$

$x=\dfrac{\pi }{12}\pm \dfrac{\pi}{12}+\dfrac{2\pi n}{3}\ ,\ n\in Z\ \ \ ili\ \ \ \ \left[\begin{array}{l}x=\dfrac{2\pi n}{3}\ ,\ n\in Z\\x=\dfrac{\pi}{6}+\dfrac{2\pi m}{3}\ ,\ m\in Z\end{array}\right\\\\\\Otvet:\ \ x_1=\dfrac{2\pi n}{3}\ ,\ \ x_2=\dfrac{\pi }{6}+\dfrac{2\pi m}{3}\ \ ,\ n,m\in Z$