Question

решите пж
не могу пожалуйста
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Answer 1

Ответ:

$\displaystyle \frac{1}{sin^2x}+\frac{1}{cos(\frac{7\pi}{2}+x)}=2\ \ ,\ \ \ ODZ:\ sinx\ne 0\ \ \to \ \ x\ne \pi n\ ,\ n\in Z\ ,\\\\\\\frac{1}{sin^2x}+\frac{1}{-sinx}-2=0\ \ ,\ \ \ \Big(\dfrac{1}{sinx}\Big)^2-\frac{1}{sinx}-2=0\ \ ,\\\\\\t=\frac{1}{sinx}\ \ ,\ \ \ t^2-t-2=0\ \ ,\ \ t_1=-1\ ,\ t_2=2\ \ (teorema\ Vieta)\\\\\\a)\ \ \frac{1}{sinx}=-1\ \ ,\ \ sinx=-1\ \ ,\ \ x=-\frac{\pi}{2}+2\pi k\ ,\ k\in Z$

$\displaystyle b)\ \ \frac{1}{sinx}=2\ \ ,\ \ sinx=\frac{1}{2}\ \ ,\ \ x=(-1)^{m}\cdot \frac{\pi}{6}+\pi m\ ,\ m\in Z\\\\\\{}\ \ \ ili\ \ \ x=\left[\begin{array}{l}\dfrac{\pi}{6}+2\pi m\ ,\ m\in Z\\\ \ \dfrac{5\pi }{6}+2\pi l\ ,\ l\in Z\end{array}\right\\\\\\Otvet:\ \ x_1=-\frac{\pi }{2}+2\pi k\ ,\ \ x_2=\dfrac{\pi}{6}+2\pi m\ ,\ \ x_3=\dfrac{5\pi }{6}+2\pi l\ ,\ k,m,l\in Z\ .\\\\\\\\P.S.\ \ m=1:\ x=\frac{\pi}{6}+2\pi \cdot 1=\frac{13\pi }{6}$