Ответ:
[tex]x \in \left( \dfrac{2\pi}{3} + 2\pi k; \ \dfrac{4\pi}{3} + 2\pi k\right), \ k \in \mathbb{Z}[/tex]
Объяснение:
[tex]2\times cos^2(x)-3 \times cos (x)-2 > 0\\\\=======\\t = cos(x)\\=======\\\\2t^2-3t-2 > 0\\\\2t^2-4t+t-2 > 0\\\\2t(t-2)+(t-2) > 0\\\\(2t+1)(t-2) > 0\\\\\begin{cases}2t+1 > 0\\t-2 > 0\end{cases}or \quad \begin{cases}2t+1 < 0\\t-2 < 0\end{cases}\\\\\\\begin{cases}2t > -1\\t > 2\end{cases} \quad or \quad \begin{cases}2t < -1\\t < 2\end{cases}\\\\\\\begin{cases}t > -\dfrac{1}{2}\\t > 2\end{cases}\quad \; or \quad \begin{cases}t < -\dfrac{1}{2}\\t < 2\end{cases}[/tex]
[tex]\bold{\Downarrow}[/tex] [tex]\bold{\Downarrow}[/tex] [tex]\bold{\Downarrow}[/tex] [tex]\bold{\Downarrow}[/tex]
[tex]t \in \left(-\infty; \ -\dfrac{1}{2}\right) \ \cup \ \left(2; \ +\infty\right)[/tex]
[tex]\boxed{t < -\dfrac{1}{2} \ \cup \ t > 2}[/tex]
[tex]cos(x) < -\dfrac{1}{2} \qquad \qquad \qquad \qquad or \qquad cos(x) > 2[/tex]
[tex]y = cos(x); \quad y= -\dfrac{1}{2}; \qquad \qquad or[/tex] Поскольку [tex]cos(x) \in [-1; \ 1][/tex] , то [tex]x \not\in \mathbb{R}[/tex]
Найдем точки
пересечения графиков:
[tex]cos(x)=-\dfrac{1}{2}[/tex]
По таблице:
[tex]x=\dfrac{2\pi}{3} \\\\x=\dfrac{4\pi}{3}\\\\\\[/tex]
Прибавим период:
[tex]x=\dfrac{2\pi}{3}+2\pi k, \ k\in \mathbb{Z} \\\\x=\dfrac{4\pi}{3}+2\pi k, \ k\in \mathbb{Z}[/tex]
По графику видим, что:
[tex]\begin{cases}x > \dfrac{2\pi}{3}+2\pi k, \ k\in \mathbb{Z} \\\\x < \dfrac{4\pi}{3}+2\pi k, \ k\in \mathbb{Z}\end{cases}[/tex]
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Answers & Comments
Ответ:
[tex]x \in \left( \dfrac{2\pi}{3} + 2\pi k; \ \dfrac{4\pi}{3} + 2\pi k\right), \ k \in \mathbb{Z}[/tex]
Объяснение:
[tex]2\times cos^2(x)-3 \times cos (x)-2 > 0\\\\=======\\t = cos(x)\\=======\\\\2t^2-3t-2 > 0\\\\2t^2-4t+t-2 > 0\\\\2t(t-2)+(t-2) > 0\\\\(2t+1)(t-2) > 0\\\\\begin{cases}2t+1 > 0\\t-2 > 0\end{cases}or \quad \begin{cases}2t+1 < 0\\t-2 < 0\end{cases}\\\\\\\begin{cases}2t > -1\\t > 2\end{cases} \quad or \quad \begin{cases}2t < -1\\t < 2\end{cases}\\\\\\\begin{cases}t > -\dfrac{1}{2}\\t > 2\end{cases}\quad \; or \quad \begin{cases}t < -\dfrac{1}{2}\\t < 2\end{cases}[/tex]
[tex]\bold{\Downarrow}[/tex] [tex]\bold{\Downarrow}[/tex] [tex]\bold{\Downarrow}[/tex] [tex]\bold{\Downarrow}[/tex]
[tex]t \in \left(-\infty; \ -\dfrac{1}{2}\right) \ \cup \ \left(2; \ +\infty\right)[/tex]
[tex]\boxed{t < -\dfrac{1}{2} \ \cup \ t > 2}[/tex]
[tex]cos(x) < -\dfrac{1}{2} \qquad \qquad \qquad \qquad or \qquad cos(x) > 2[/tex]
[tex]y = cos(x); \quad y= -\dfrac{1}{2}; \qquad \qquad or[/tex] Поскольку [tex]cos(x) \in [-1; \ 1][/tex] , то [tex]x \not\in \mathbb{R}[/tex]
Найдем точки
пересечения графиков:
[tex]cos(x)=-\dfrac{1}{2}[/tex]
По таблице:
[tex]x=\dfrac{2\pi}{3} \\\\x=\dfrac{4\pi}{3}\\\\\\[/tex]
Прибавим период:
[tex]x=\dfrac{2\pi}{3}+2\pi k, \ k\in \mathbb{Z} \\\\x=\dfrac{4\pi}{3}+2\pi k, \ k\in \mathbb{Z}[/tex]
По графику видим, что:
[tex]\begin{cases}x > \dfrac{2\pi}{3}+2\pi k, \ k\in \mathbb{Z} \\\\x < \dfrac{4\pi}{3}+2\pi k, \ k\in \mathbb{Z}\end{cases}[/tex]
[tex]\bold{\Downarrow}[/tex] [tex]\bold{\Downarrow}[/tex] [tex]\bold{\Downarrow}[/tex] [tex]\bold{\Downarrow}[/tex]
[tex]x \in \left( \dfrac{2\pi}{3} + 2\pi k; \ \dfrac{4\pi}{3} + 2\pi k\right), \ k \in \mathbb{Z}[/tex]