Question

Помогите решить, пожалуйста ❤️. алгебра 10 класс

Answer 1

$2\, cos^2x+\sqrt2\, sinx>2\\\\2(1-sin^2x)+\sqrt2\, sinx-2>0\\\\2-2\, sin^2x+\sqrt2\, sinx-2>0\\\\2sin^2x-\sqrt2\, sinx<0\\\\\sqrt2\, sinx\cdot (\sqrt2\, sinx-1)<0\\\\t=sinx\; ,\; |t|\leq 1\; \; ,\; \; \sqrt2\, t\cdot (\sqrt2\, t-1)<0\; \; ,\; \; +++(0)---(\frac{1}{\sqrt2} )+++\\\\0<t<\frac{1}{\sqrt2} \; \; \; \Rightarrow \; \; 0<sinx<\frac{\sqrt2}{2}\; \; \Rightarrow \; \; \left \{ {{sinx>0} \atop {sinx<\frac{\sqrt2}{2}}} \right.$

$\left \{ {{2\pi n<x<\pi +2\pi n\; ,\; n\in Z\; \quad } \atop {-\frac{5\pi }{4}+2\pi k<x<\frac{\pi }{4}+2\pi k\; ,\; k\in Z}} \right. \; \; \Rightarrow \\\\x\in (2\pi n;\frac{\pi}{4}+2\pi n)\cup (\frac{3\pi }{4}+2\pi n,\; \pi +2\pi n)\; ,\; n\in Z$

Answer 2

$2 {cos}^{2} x + \sqrt{2} sinx > 2 \\ 2 - 2 {sin}^{2} x + \sqrt{2}sin x - 2 > 0 \\ 2 {sin}^{2} x - \sqrt{2}sin < 0 \\ \sqrt{2} sinx( \sqrt{2} sinx - 1) < 0 \\ 0 < sinx < \frac{ \sqrt{2} }{2} \\ \\2\pi \times n < x < \frac{\pi}{4} + 2\pi \times n$

3пи/4 + пи*n <х< пи + пи*n