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Ответ:

[tex]\displaystyle 1)\ \ sin^3x(1+ctgx)+cos^2x(1+tgx)=2\sqrt{sinx\cdot cosx}\ ,\\\\ODZ:\ x\ne \dfrac{\pi n}{2}\ \ \ i\ \ \ \ sinx\cdot cosx\geq 0\ \ \Rightarrow x\in (2\pi l\ ;\ \dfrac{\pi}{2}+2\pi l)\cup (\pi +2\pi l\ ;\ \dfrac{3\pi }{2}+2\pi l\ )\\\\sin^3x\Big(1+\frac{cosx}{sinx}\Big)+cos^3x\Big(1+\frac{sinx}{cosx}\Big)=2\sqrt{sinx\cdot cosx}\\\\sin^3x\cdot \frac{sinx+cosx}{sinx}+cos^3x\cdot \frac{sinx+cosx}{cosx}=2\sqrt{sinx\cdot cosx}[/tex]

[tex]sin^2x\cdot (sinx+cosx)+cos^2x(sinx+cosx)=2\sqrt{sinx\cdot cosx}\\\\(sinx+cosx)\cdot (\underbrace{sin^2x+cos^2x}_{1})=2\sqrt{sinx\cdot cosx}[/tex]  

[tex]\displaystyle 2\sqrt{sinx\cdot cosx}=sinx+cosx\ \ \Rightarrow \ \ \left\{\begin{array}{l}4sinx\cdot cosx=(sinx+cosx)^2\\sinx+cosx\geq 0\end{array}\right\\\\\\\left\{\begin{array}{l}4sinx\cdot cosx=sin^2x+cos^2x+2sinx\cdot cosx\\sinx+cosx\geq 0\end{array}\right\ \ \left\{\begin{array}{l}sin^2x+cos^2x-2sinx\cdot cosx=0\\sinx+cosx\geq 0\end{array}\right[/tex]

[tex]\left\{\begin{array}{l}(sinx-cosx)^2=0\\sinx+cosx\geq 0\ |:cosx\ne 0\end{array}\right\ \ \left\{\begin{array}{l}sinx-cosx=0\\tgx\geq -1\end{array}\right\ \ \left\{\begin{array}{l}tgx=1\\tgx\geq -1\end{array}\right\\\\\\\left\{\begin{array}{l}x=\dfrac{\pi}{4}+\pi n\\-\dfrac{\pi}{4}+\pi k\leq x < \dfrac{\pi }{2}+\pi k\end{array}\right\ \ \Rightarrow \ \ x=\dfrac{\pi }{4}+\pi n\ ,\ n\in \mathbb{Z}\\\\\\Otvet:\ x=\dfrac{\pi }{4}+\pi n\ ,\ n\in \mathbb{Z}\ .[/tex]  

[tex]2)\ \ sin2x+5(sinx+cosx)=-1[/tex]

Замена:  [tex]t=sinx+cosx\ \ \to \ \ t^2=\underbrace{sin^2x+cos^2x}_{1}+2sinx\cdot cosx=1+sin2x\ ,\\\\sin2x=t^2-1\\\\(t^2-1)+5t=-1\ \ \to \ \ \ t^2+5t=0\ \ ,\ \ t\cdot (t+5)=0\ \ ,\\\\a)\ \ t_1=0\ ,\ sinx+cosx=0\ \Big|:cosx\ne 0\ \ ,\ \ \ tgx=-1\ ,\ x=-\dfrac{\pi}{4}+\pi n\ ,\ n\in \mathbb{Z}\\\\b)\ \ t_2=-5\ \ ,\ \ sinx+cosx=-5\ \ \Rightarrow \ \ \ x\in \varnothing\\\\{}\ \ \ \ \ \ \Big[\ sinx+cosx=\sqrt2cos(x-\dfrac{\pi }{4})\ \ ,\ \ -\sqrt2\leq \sqrt2cos(x-\dfrac{\pi }{4})\leq \sqrt2\ \Big][/tex]

Уравнение не имеет решений, так как минимальное значение, которое может принимать сумма sinx и cosx , не может равняться  -5 .

 Ответ:  [tex]x=-\dfrac{\pi}{4}+\pi n\ ,\ n\in \mathbb{Z}[/tex]  .  

[tex]3)\ \ cosx+sinx-\sqrt{1-2cos^2x}=0\\\\cosx+sinx-\sqrt{-(2cos^2x-1)}=0\\\\sinx+cosx-\sqrt{-cos2x}=0\ \ ,\ \ \ \sqrt{-cos2x}=sinx+cosx\ \ \Rightarrow[/tex]  

[tex]\left\{\begin{array}{l}-cos2x=(sinx+cosx)^2\\sinx+cosx\geq 0\ \Big|:cosx\ne 0\end{array}\right\ \ \left\{\begin{array}{l}-cos2x=1+sin2x\\tgx\geq -1\end{array}\right\\\\\\\left\{\begin{array}{l}sin2x+cos2x=-1\\-\dfrac{\pi}{4}+\pi k\leq x < \dfrac{\pi}{2}+\pi k\end{array}\right[/tex]

Решим первое уравнение системы

[tex]sin2x+cos2x=-1\ \Big|:\sqrt2\ \ \ \Rightarrow \ \ \ \dfrac{1}{\sqrt2}\, sin2x+\dfrac{1}{\sqrt2}\, cos2x=-\dfrac{1}{\sqrt2}\\\\sin\Big(2x+\dfrac{\pi}{4}\Big)=-\dfrac{1}{\sqrt2}\\\\\\2x+\dfrac{\pi}{4}=-\dfrac{\pi }{4}+2\pi m\ \ \ \ ili\ \ \ \ 2x+\dfrac{\pi}{4}=-\dfrac{3\pi }{4}+2\pi m\ \ ,\ m\in \nathbb{Z}\\\\x=-\dfrac{\pi}{4}+\pi m\qqiuad \qquad \qquad \ ili\qquad x=-\dfrac{\pi}{2}+\pi m\ \ ,\ m\in \mathbb{Z}[/tex]

[tex]\left[\begin{array}{l}\left\{\begin{array}{l}x=-\dfrac{\pi}{4}+\pi m\ ,\ m\in Z\\-\dfrac{\pi}{4}+\pi k\leq x < \dfrac{\pi}{2}+\pi k\ ,\ k\in Z\end{array}\right\\\left\{\begin{array}{l}x=-\dfrac{\pi}{2}+\pi m\ ,\ m\in Z\\-\dfrac{\pi}{4}+\pi k\leq x < \dfrac{\pi}{2}+\pi k\ ,\ k\in Z\end{array}\right\end{array}\right\ \ \ \ \Rightarrow \\\\\\Otvet:\ x=-\dfrac{\pi}{4}+\pi m\ ,\ m\in Z\ \ ,\ m\in Z\ .[/tex]