Question

cos(x)+sqrt((2-sqrt2)/2*(sinx+1))=0

Answer 1

$\displaystyle cosx+\sqrt{\frac{2-\sqrt{2}}{2}*(sinx+1)}=0$

$\displaystyle \sqrt{\frac{2-\sqrt{2}}{2}(sinx+1)}=-cosx$

решение данного уравнения возможно только при условии

cosx ≤0

При условии что cos x ≤0 возведем обе части уравнения в квадрат

$\displaystyle (\frac{2-\sqrt{2}}{2})(sinx+1)=cos^2x\\\\(2-\sqrt{2})(sinx+1)=2(1-sin^2x)\\\\(2-\sqrt{2})sinx+2-\sqrt{2}=2-2sin^2x\\\\2sin^2x+(2-\sqrt{2})*sinx-\sqrt{2}=0\\\\sinx=t; |t|\leq 1\\\\2t^2+(2-\sqrt{2})t-\sqrt{2}=0$

$\displaystyle D=(2-\sqrt{2})^2-4*2*(-\sqrt{2})=4-4\sqrt{2}+2+8\sqrt{2}=6+4\sqrt{2}=(2+\sqrt{2})^2\\\\t_{1.2}=\frac{-(2-\sqrt{2})\pm(2+\sqrt{2})}{4}\\\\t_1=-1; t_2=\frac{\sqrt{2}}{2}$

$\displaystyle sinx=-1; \\\\x=-\frac{\pi}{2}+2 \pi n; n \in Z \\\\sinx=\frac{\sqrt{2}}{2}\\\\ x=\frac{\pi}{4}+2 \pi n; n \in Z; x= \frac{3\pi}{4}+2 \pi n; n\in Z$

Проверим условие cosx≤0

π/2+2πn≤x≤3π/2+2πn; n∈Z

Окончательный ответ:

x= -π/2+ 2πn; n∈Z

x= 3π/4+2πn; n∈Z