Question

алгебра: решите системы тригонометрических уравнений: {x+y=5/6π, cos^2x+cos^2y=1/4; {x+y=2/3π, 2cosx+4cosy=3:​

Answer 1

Ответ:

(2П/3+2пk; 2Пl)  k=-l  k,l ∈Z

(П/3+2Пk; П/3+2Пl);

Объяснение:

x+y=П-П/3     x=П-(П/3+y)

2cosx+4cosy=3  

2cos(П-(П/3+y))+4cosy=3

-2cos(П/3+y)+4cosy=3

-cosy+√3siny+4cosy=3

√3cosy+siny=√3 :2

cos(y-П/6)=√3/2

y-П/6=П/6+2Пk  y=П/3+2Пk  x=П-(П/3+П/3+2Пk)

y-П/6=-П/6+2Пk  y=2Пk  x=П-(П/3+2Пk)=2П/3-2пk

Answer 2

Ответ:

1) $(x,y)=\left(\dfrac{\pi}{2}+\pi k,\dfrac{\pi}{3}-\pi k\right),\left(\dfrac{\pi}{3}+\pi k,\dfrac{\pi}{2}-\pi k\right),k\in\mathbb{Z}$

2) $(x,y)=\left(\dfrac{\pi}{3}+2\pi k,\dfrac{\pi}{3}-2\pi k\right),\left(\dfrac{2\pi}{3}+2\pi k,-2\pi k\right), k\in\mathbb{Z}$

Объяснение:

1)

$\begin{cases}x+y=\dfrac{5\pi}{6},\\ \cos^2{x}+\cos^2{y}=\dfrac{1}{4}\end{cases}\Leftrightarrow \begin{cases}y=\dfrac{5\pi}{6}-x,\\ \cos^2{x}+\cos^2{\left(\dfrac{5\pi}{6}-x\right)}=\dfrac{1}{4}\end{cases}\Leftrightarrow\\\Leftrightarrow \begin{cases}y=\dfrac{5\pi}{6}-x,\\ \cos^2{x}+\left(-\dfrac{\sqrt{3}\cos{x}}{2}+\dfrac{\sin{x}}{2}\right)^2=\dfrac{1}{4}\end{cases}\Leftrightarrow$

$\Leftrightarrow \begin{cases}y=\dfrac{5\pi}{6}-x,\\ \cos^2{x}+\dfrac{\sin^2{x}-2\sqrt{3}\sin{x}\cos{x}+3\cos^2{x}}{4}}=\dfrac{1}{4}\end{cases}\Leftrightarrow\\\Leftrightarrow \begin{cases}y=\dfrac{5\pi}{6}-x,\\ \cos^2{x}+\dfrac{\sin^2{x}+\cos^2{x}-2\sqrt{3}\sin{x}\cos{x}+2\cos^2{x}}{4}}=\dfrac{1}{4}\end{cases}\Leftrightarrow\\\Leftrightarrow \begin{cases}y=\dfrac{5\pi}{6}-x,\\ \dfrac{1-2\sqrt{3}\sin{x}\cos{x}+6\cos^2{x}}{4}}=\dfrac{1}{4}\end{cases}\Leftrightarrow$

$\Leftrightarrow \begin{cases}y=\dfrac{5\pi}{6}-x,\\ 1-2\sqrt{3}\sin{x}\cos{x}+6\cos^2{x}}=1\end{cases}\Leftrightarrow \begin{cases}y=\dfrac{5\pi}{6}-x,\\ 6\cos^2{x}-2\sqrt{3}\sin{x}\cos{x}}=0\end{cases}\Leftrightarrow\\\Leftrightarrow \begin{cases}y=\dfrac{5\pi}{6}-x,\\ \cos{x}(\sqrt{3}\cos{x}-\sin{x})}=0\end{cases}\Leftrightarrow \begin{cases}y=\dfrac{5\pi}{6}-x,\\ \left [ {{\cos{x}=0,} \atop {\sqrt{3}\cos{x}=\sin{x}}} \right. \end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}y=\dfrac{5\pi}{6}-x,\\ \left [ {{\cos{x}=0,} \atop {tgx=\sqrt{3}}} \right. \end{cases}\Leftrightarrow \begin{cases}\cos{x}=0,\\ y=\dfrac{5\pi}{6}-x \end{cases} or \begin{cases}tgx=\sqrt{3},\\ y=\dfrac{5\pi}{6}-x \end{cases}\Leftrightarrow\\\Leftrightarrow \begin{cases}x=\dfrac{\pi}{2}+\pi k,\\ y=\dfrac{5\pi}{6}-\left(\dfrac{\pi}{2}+\pi k\right) \end{cases} or \begin{cases}x=\dfrac{\pi}{3}+\pi k,\\ y=\dfrac{5\pi}{6}-\left(\dfrac{\pi}{3}+\pi k\right)\end{cases} (k\in\mathbb{Z})$

$\Leftrightarrow \begin{cases}x=\dfrac{\pi}{2}+\pi k,\\ y=\dfrac{\pi}{3}-\pi k\end{cases} or\begin{cases}x=\dfrac{\pi}{3}+\pi k,\\ y=\dfrac{\pi}{2}-\pi k\end{cases} (k\in\mathbb{Z})$

2)

$\begin{cases}x+y=\dfrac{2\pi}{3},\\ 2\cos{x}+4\cos{y}=3\end{cases}\Leftrightarrow \begin{cases}y=\dfrac{2\pi}{3}-x,\\ 2\cos{x}+4\cos{\left(\dfrac{2\pi}{3}-x\right)}=3\end{cases}\Leftrightarrow\\\Leftrightarrow \begin{cases}y=\dfrac{2\pi}{3}-x,\\ 2\cos{x}+4\cdot\left(-\dfrac{\cos{x}}{2}+\dfrac{\sqrt{3}\sin{x}}{2}\right)=3\end{cases}\Leftrightarrow\\\Leftrightarrow \begin{cases}y=\dfrac{2\pi}{3}-x,\\ 2\cos{x}-2\cos{x}+2\sqrt{3}\sin{x}=3\end{cases}\Leftrightarrow$

$\Leftrightarrow \begin{cases}y=\dfrac{2\pi}{3}-x,\\ \sin{x}=\dfrac{\sqrt{3}}{2}\end{cases}\Leftrightarrow \begin{cases}y=\dfrac{2\pi}{3}-\left(\dfrac{\pi}{3}+2\pi k\right),\\ x=\dfrac{\pi}{3}+2\pi k\end{cases}or\\ \begin{cases}y=\dfrac{2\pi}{3}-\left(\dfrac{2\pi}{3}+2\pi k\right),\\ x=\dfrac{2\pi}{3}+2\pi k\end{cases}(k\in\mathbb{Z})\Leftrightarrow\\\Leftrightarrow \begin{cases}x=\dfrac{\pi}{3}+2\pi k,\\ y=\dfrac{\pi}{3}-2\pi k\end{cases} or\begin{cases}x=\dfrac{2\pi}{3}+2\pi k,\\ y=-2\pi k\end{cases}$$(k\in\mathbb{Z})$