-2sin²x-7sinx+3=0 + Найти корни, принадлежащие к cosx≤0. Created by mummyandi1. algebra-ru

-2sin²x-7sinx+3=0 + Найти корни, принадлежащие к cosx≤0...

igundane

$-2sin^2x-7sinx+3=0\Leftrightarrow 2sin^2x+7sinx-3=0\\sin^2x+\frac{7}{2}sinx+\frac{49}{16}=\frac{73}{16}\Leftrightarrow \left ( sinx+\frac{7}{4} \right )^2=\frac{73}{16}\\sinx=-\frac{7}{4}\pm \frac{\sqrt{73}}{4}\Rightarrow sinx=-\frac{7}{4}+\frac{\sqrt{73}}{4}\\cosx\leq 0\Rightarrow x=\pi-arcsin\left ( \frac{-7+\sqrt{73}}{4} \right )+2\pi k,k\in \mathbb{Z}$

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stanmat

Ответ:

$x = \pi - \arcsin \frac{{7 - \sqrt {73} }}{{ - 4}} + 2\pi k,k \in {\bf{Z}}$

Объяснение:

$\begin{array}{l} - 2{\sin ^2}x - 7\sin x + 3 = 0\\\sin x = \frac{{7 \pm \sqrt {49 + 4 \cdot 2 \cdot 3} }}{{ - 2 \cdot 2}} = \frac{{7 \pm \sqrt {73} }}{{ - 4}}\end{array}$

Т.к. |sinx|<=1, то $\sin x = \frac{{7 - \sqrt {73} }}{{ - 4}}$

$\begin{array}{l}\sin x = \frac{{7 - \sqrt {73} }}{{ - 4}}\\\left[ \begin{array}{l}x = \arcsin \frac{{7 - \sqrt {73} }}{{ - 4}} + 2\pi k\\x = \pi - \arcsin \frac{{7 - \sqrt {73} }}{{ - 4}} + 2\pi k\end{array} \right.\\x = \pi - \arcsin \frac{{7 - \sqrt {73} }}{{ - 4}} + 2\pi k,k \in {\bf{Z}}\end{array}$

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