Question

Cos2x+sin^2x=0 на промежутке (0; 3п)

Answer 1

cos2x + sin²x = 0    на промежутке (0; 3π)

$cos~2x+sin^2x=0\\\\ cos^2x-sin^2x+sin^2x=0\\ \\ cos^2x=0~~~\Leftrightarrow~~~ cos~x=0\\ \\ \boldsymbol{x=\dfrac{\pi }{2}+\pi n,~~n\in Z}$

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$n=-1;~~~x_0=\dfrac{\pi }{2}-\pi=-0,5\pi;~~~x_0\notin (0;3\pi )\\ \\ n=0;~~~x_1=\dfrac{\pi }{2}=0,5\pi;~~~x_1\in (0;3\pi )\\ \\ n=1;~~~x_2=\dfrac{\pi }{2}+\pi=1,5\pi;~~~x_2\in (0;3\pi )\\ \\ n=2;~~~x_3=\dfrac{\pi }{2}+2\pi=2,5\pi;~~~x_2\in (0;3\pi )\\ \\n=3;~~~x_4=\dfrac{\pi }{2}+3\pi=3,5\pi;~~~x_4\notin (0;3\pi )\\ \\ \\ \boxed{\boldsymbol{~~x_1=0,5\pi;~~~x_2=1,5\pi;~~~x_3=2,5\pi ~~}}$