Question

Решите уравнение, пожалуйста! Желаьельно подробнее) Заранее благодарна

Answer 1

Ответ:

решение на фотографии

Answer 2

Ответ:

$2sin^2x+(2-\sqrt2)\, cosx+\sqrt2-2=0\ \ ,\ \ x\in \Big[\; \dfrac{5\pi}{2}\, ;\, \dfrac{7\pi}{2}\, \Big]\\\\\\2(1-cos^2x)+(2-\sqrt2)\, cosx+(\sqrt2-2)=0\\\\2-2cos^2x+(2-\sqrt2)\, cosx+(\sqrt2-2)=0\\\\2cos^2x-(2-\sqrt2)\, cosx-\sqrt2=0\\\\D=(2-\sqrt2)^2+4\cdot 2\sqrt2=4-4\sqrt2+2+8\sqrt2=4+4\sqrt2+2=(2+\sqrt2)^2\\\\a)\ \ cosx=\dfrac{(2-\sqrt2)-(2+\sqrt2)}{4}=\dfrac{-2\sqrt2}{4}=-\dfrac{\sqrt2}{2}\\\\x=\pm (\pi -\dfrac{\pi}{4})+2\pi n\ ,\ \ x=\pm \dfrac{3\pi}{4}+2\pi n\ ,\ n\in Z$

$cosx=\dfrac{(2-\sqrt2)+(2+\sqrt2)}{4}=\dfrac{4}{4}=1\\\\x=2\pi k\ ,\ k\in Z\\\\c)\ \ x\in \Big[\ \dfrac{5\pi}{2}\, ;\, \dfrac{7\pi }{2}\ \Big]\\\\\dfrac{5\pi}{2}\leq \dfrac{3\pi}{4}+2\pi n\leq \dfrac{7\pi}{2}\ \ ,\ \ \dfrac{7}{4}\leq 2n\leq \dfrac{11}{4}\ \,\ \ \ \dfrac{7}{8}\leq n\leq \dfrac{11}{8}\ \ ,\ \ \ \dfrac{7}{8}\leq x\leq 1\dfrac{3}{8}\ \to \\\\\\n=1:\ \ x_1=\dfrac{3\pi}{4}+2\pi\cdot 1=\dfrac{11\pi}{8}$

$\dfrac{5\pi}{2}\leq -\dfrac{3\pi}{4}+2\pi n\leq \dfrac{7\pi}{2}\ \ ,\ \ \dfrac{13}{4}\leq 2n\leq \dfrac{17}{4}\ \,\ \ \ \dfrac{13}{8}\leq n\leq \dfrac{17}{8}\ \ ,\ \ \ 1\dfrac{5}{8}\leq x\leq 2\dfrac{1}{8}\ \to \\\\\\n=2:\ \ x_2=-\dfrac{3\pi}{4}+2\pi\cdot 2=\dfrac{13\pi}{4}\\\\\\2\pi k\notin \Big[\ \dfrac{5\pi}{2}\, ;\, \dfrac{7\pi}{2}\ \Big]\\\\Otvet:\ \ 1)\ x_1=\pm \dfrac{3\pi}{4}+2\pi n\ ,\ \ x_2=2\pi k\ ,\ \ n,k\in Z\ ;$

             ${}\qquad \qquad 2)\ \dfrac{11\pi}{4}\ ,\ \dfrac{13\pi}{4}\in \Big[\ \dfrac{5\pi }{2}\, ;\, \dfrac{7\pi }{2}\ \Big]\ .$