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Задание 7179
A) $$3*2^{\cos x+3\sqrt{1-\sin ^{2}x}}+11 *2^{2 \cos x}-34=0\Leftrightarrow$$ $$3*2^{\cos x+3\sqrt{\cos ^{2}x}}+11*2^{2 \cos x}-34=0\Leftrightarrow$$ $$3*2^{\cos x+3\left | \cos ^{2}x \right |}+11*2^{2 \cos x}-34=0$$
1) при $$\cos x\geq 0$$$$\Leftrightarrow$$ $$x \in [-\frac{\pi}{2}+2 \pi n, \frac{\pi}{2}+2 \pi n], n \in Z$$: $$3*2^{4 \cos x}+11*2^{2 \cos x}-34=0$$
Пусть $$2^{2 \cos x}=y>0$$, тогда $$3y^{2}+11y-34=0$$: $$D=121+408=529$$
$$\left[\begin{matrix}y_{1}=\frac{-11+23}{6}=2\\y_{2}=\frac{-11-23}{6}<0\end{matrix}\right.\Leftrightarrow$$ $$2 ^{2 \cos x}=2\Leftrightarrow$$ $$2 \cos x=1\Leftrightarrow$$ $$\cos x=\frac{1}{2}\Leftrightarrow$$ $$x=\pm \frac{\pi}{3}+2 \pi n , n \in Z$$
2) при $$\cos x<0$$: $$3*2^{\cos x-3 \cos x}+11*2^{2 \cos x}-34=0\Leftrightarrow$$$$3*2^{-2\cos x}+11*2^{2 \cos x}-34=0$$
Пусть $$2^{2 \cos x}=y>0$$ , тогда $$\frac{3}{y}+11*y-34=0\Leftrightarrow$$ $$\frac{11y^{2}-34y+3}{y}=0\Leftrightarrow$$ $$11y^{2}-34y+3=0$$
$$D=1156-132=1024$$
$$\left[\begin{matrix}y_{1}=\frac{34+32}{22}=3\\y_{2}=\frac{34-32}{22}=\frac{1}{11}\end{matrix}\right.\Leftrightarrow$$ $$\left[\begin{matrix}2^{2 \cos x}=3\\2 ^{2 \cos x}=\frac{1}{11}\end{matrix}\right.\Leftrightarrow$$ $$\left[\begin{matrix}4^{\cos x}=3\\4^{\cos x}=\frac{1}{11}\end{matrix}\right.\Leftrightarrow$$ $$\left[\begin{matrix}\cos x=\log_{4}3>0\Rightarrow \varnothing\\\cos x=\log_{4}\frac{1}{11}<-1\Rightarrow \varnothing & &\end{matrix}\right.$$
Б) На промежутке $$[-\frac{\pi}{2};\frac{5\pi}{2}]$$:
$$\frac{\pi}{3}+2 \pi n$$: $$\frac{\pi}{3};\frac{7\pi}{3}$$
$$-\frac{\pi}{3}+2 \pi n$$: $$\frac{\pi}{3};\frac{5\pi}{3}$$
Задание 7199
A) Воспользуемся формулами приведения: $$\sin (2x+\frac{5 \pi}{2})=\sin (\frac{5 \pi}{2}+2x)=$$$$\sin (\frac{\pi}{2}+2x)=\cos 2x$$; $$\cos (x-\frac{7 \pi}{2})=\cos (\frac{7 \pi}{2}-x)=$$$$\cos (\frac{3 \pi}{2}-x)=-\sin x$$
Тогда получим: $$\cos 2x+3 \sin x-1-2 \sin x=0\Leftrightarrow$$ $$1-2 \sin ^{2}x+\sin x-1=0\Leftrightarrow$$ $$\sin x-2 \sin ^{2}x=0\Leftrightarrow$$ $$\sin x(1-2 \sin x)=0\Leftrightarrow$$ $$\left[\begin{matrix}\sin x=0\\\sin x=\frac{1}{2}\end{matrix}\right.\Leftrightarrow$$$$\left[\begin{matrix}x=\pi n, n \in Z\\x=\frac{\pi}{6}+2 \pi k\\x=\frac{5 \pi}{6}+ 2 \pi k\end{matrix}\right.$$
Б) Найдем корни, принадлежащие $$[-\frac{3 \pi}{2}; \pi]$$:
$$\frac{5 \pi}{6}+2 \pi k$$:$$ -\pi-\frac{\pi}{6}=-\frac{7 \pi}{6}$$; $$\pi-\frac{\pi}{6}=\frac{5 \pi}{6}$$
$$\pi n$$: $$-\pi ;0; \pi$$.
$$\frac{\pi}{6}+\pi k$$: $$0+\frac{\pi}{6}=\frac{\pi}{6}$$
Задание 7220
A) ОДЗ: $$\left\{\begin{matrix}tg (x+\frac{\pi}{6})\neq 0\\\sin x\neq 0\\\cos (x+\frac{\pi}{6})\neq 0\end{matrix}\right.\Leftrightarrow$$$$\left\{\begin{matrix}x+\frac{\pi}{6}\neq \pi n , n \in Z\\x\neq \pi k, k \in Z\\x+\frac{\pi}{6}\neq \frac{\pi}{2}+\pi m, m \in Z\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}x\neq -\frac{\pi}{6}\\x\neq \pi k\\x\neq \frac{\pi}{3}+\pi m, n,k,m \in Z\end{matrix}\right.$$
Решение: $$ctg \frac{110}{6}=-\sqrt{3}$$; $$tg(x+\frac{\pi}{6})=\frac{\sin (x+\frac{\pi}{6})}{\cos (x+\frac{\pi}{6})}=$$$$\frac{\frac{\sqrt{3}}{2}\sin x +\frac{1}{2} \cos x}{\frac{\sqrt{3}}{2} \cos x -\frac{1}{2} \sin x}=$$$$\frac{\sqrt{3} \sin x+\cos x}{\sqrt{3} \cos x-\sin x}$$; $$ctg x=\frac{\cos x}{\sin x}$$;
Получим $$-\sqrt{3} (\frac{\sqrt{3} \sin x+\cos x}{\sqrt{3} \cos x -\sin x})=$$$$\frac{2 \cos x}{\sin x}+3\Leftrightarrow$$ $$\frac{-3 \sin x -\sqrt{3} \cos x}{\sqrt{3} \cos x-\sin x}=$$$$\frac{2 \cos x+3 \sin x}{\sin x}\Leftrightarrow$$ $$-3\sin ^{2}x-\sqrt{3} \sin x\cos x=$$$$2\sqrt{3}\cos ^{2}x+3\sqrt{3} \cos x \sin x-2 \sin x \cos x-3 \sin ^{2}x\Leftrightarrow$$ $$2\sqrt{3} \cos ^{2}x+4\sqrt{3} \sin x \cos x-2 \sin x \cos x=0\Leftrightarrow$$ $$2 \cos x(\sqrt{3} \cos x+\sin x(2\sqrt{3}-1))=0\Leftrightarrow$$ $$\left[\begin{matrix}\cos x=0\\\sin x (2\sqrt{3} -1)+\sqrt{3} \cos x=0\end{matrix}\right.\Leftrightarrow$$$$\left[\begin{matrix}x=\frac{\pi}{2} +\pi n\\tg x=\frac{-3}{2\sqrt{3}-1}\end{matrix}\right.\Leftrightarrow$$ $$\left[\begin{matrix}x=\frac{\pi}{2}+ \pi n\\x=-arctg \frac{\sqrt{3}}{2\sqrt{3}-1}+ \pi k , n,k \in Z\end{matrix}\right.$$
Б) С учетом тригонометрической окружности : $$\frac{\pi}{2} +\pi n$$ :$$-\frac{3 \pi}{2}$$; $$-\frac{\pi}{2}$$ ;$$\frac{\pi}{2}$$; $$\frac{3 \pi}{2}$$
$$-arctg \frac{\sqrt{3}}{2\sqrt{3}-1}+\pi k$$: $$-\pi-arctg \frac{\sqrt{3}}{2\sqrt{3}-1}$$;$$-arctg \frac{\sqrt{3}}{2\sqrt{3}-1};$$ $$\pi-arctg \frac{\sqrt{3}}{2\sqrt{3}-1}$$.
Задание 7322
А) Учтем, что: $$\sin \frac{x}{2}+\sin \frac{3x}{2}=$$$$2\sin \frac{\frac{x}{2}+\frac{3x}{2}}{2}\cos \frac{\frac{x}{2}-\frac{3x}{2}}{2}=$$$$2 \sin x \cos x$$
Выразим: $$2 \sin x cos \frac{x}{2}=$$$$\sin (-x)\Leftrightarrow$$ $$2 \sin x \cos \frac{x}{2}+\sin x=0\Rightarrow$$ $$\sin x(2\cos \frac{x}{2}+1)=0\Leftrightarrow$$ $$\left[\begin{matrix}\sin x=0\\2 \cos \frac{x}{2}+1=0\end{matrix}\right.\Leftrightarrow$$$$\left[\begin{matrix}\sin x=0\\\cos \frac{x}{2}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow$$ $$\left[\begin{matrix}x=\pi n , n \in Z\\\frac{x}{2}=\pm \frac{2\pi}{3}+2 \pi k,k \in Z\end{matrix}\right.\Leftrightarrow$$ $$\left[\begin{matrix}x=\pi n,n\in Z\\x=\pm \frac{4 \pi}{3}+4 \pi k, k \in Z\end{matrix}\right.$$
ОДЗ: $$\left\{\begin{matrix}\sin (-x)>0\\\sin (-x)\neq 1\\2 \sin x \cos \frac{x}{2}>0\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}\sin x <0\\\sin x \neq -1\\\sin x \cos \frac{x}{2}>0\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}x \in (-\pi +2 \pi n , 2 \pi n) (2)\\x \neq -\frac{\pi}{2}+2 \pi n \\\sin x \cos \frac{x}{2}>0 (1)\end{matrix}\right.$$
С учетом (2) $$x =\pi n$$ не подходит, $$x=-\frac{4 \pi}{3} +4 \pi n$$ не подходит. Подставим $$x= \frac{4 \pi}{3} + 4 \pi k$$ в (1) : $$\sin (\frac{4 \pi}{3})\cos \frac{\frac{4\pi}{3}}{2}=$$$$-\frac{\sqrt{3}}{2}\cos \frac{ \pi}{3}=$$$$-\frac{\sqrt{3}}{2}*(-\frac{1}{2})>0$$$$\Rightarrow$$ $$\frac{4 \pi}{3}+2 \pi k, k \in Z$$-корень
Б) На промежутке $$[-2 \pi ; 2 \pi]$$: $$-2 \pi\leq \frac{4 \pi}{3}+2 pi k \leq 2 \pi\Leftrightarrow$$ $$-\frac{20 \pi}{3}\leq 4 \pi\leq k\leq \frac{2 \pi}{3}\Leftrightarrow$$ $$-\frac{10}{12}\leq k\leq \frac{1}{6}\Rightarrow$$ $$k=0\Rightarrow$$ $$\frac{4 \pi}{3}+0*\pi =\frac{4 \pi}{3}$$
Задание 7440
Задание 7862
а) Решите уравнение: $$\sin(\frac{\pi}{3}-2x)=-2\cos^{2}(\frac{\pi}{12}+x)-1$$;
б) Укажите корни этого уранения, принадлежащие отрезку $$\begin{bmatrix}\frac{\pi}{2}&;\frac{7\pi}{2}\end{bmatrix}$$
a) $$\sin(\frac{\pi}{3}-2x)=-2\cos^{2}(\frac{\pi}{12}+x)-1$$
$$\sin(\frac{\pi}{3}-2x)=-(2\cos^{2}(\frac{\pi}{12}+x)-1)-2$$
$$\sin(\frac{\pi}{3}-2x)=-\cos(\frac{\pi}{6}+2x)-2$$
Заметим, что : $$\cos(\frac{\pi}{6}+2x)=\cos(\frac{\pi}{2}-(\frac{\pi}{3}-2x))=\sin(\frac{\pi}{3}-2x)$$
$$\sin(\frac{\pi}{3}-2x)=-\sin(\frac{\pi}{3}-2x)-2$$
$$\sin(\frac{\pi}{3}-2x)=-2$$ $$\Rightarrow$$ $$\sin(\frac{\pi}{3}-2x)=-1$$ $$\Rightarrow$$ $$\frac{\pi}{3}-2x=-\frac{\pi}{2}+2\pi k$$ $$\Rightarrow$$ $$-2x=-\frac{5\pi}{6}+2\pi k$$ $$\Rightarrow$$ $$x=\frac{5\pi}{12}+\pi k$$, $$k\in Z$$
б) с помощью двойного неравенства отберем корни: $$\frac{\pi}{2}\leq \frac{5\pi}{12}+\pi k \leq \frac{7\pi}{2}\Leftrightarrow$$$$\frac{\pi}{12}\leq \pi k \leq \frac{37\pi}{12}\Leftrightarrow$$$$\frac{1}{12}\leq k\leq \frac{37}{12}$$.
Тогда $$k=1: x=\frac{5\pi}{12}$$; $$k=2: x=\frac{17\pi}{12}$$; $$k=2: x=\frac{29\pi}{12}$$