3-4 cos^2x>0 как решить неравенство?. Created by ladyveronica. matematika-ru

1) 2) объединим промежутки

3-4 cos^2x>0 как решить неравенство?...

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$\displaystyle 3-4cos^2x\ \textgreater \ 0$

$\displaystyle 3\ \textgreater \ 4cos^2x$

$\displaystyle cos^2x\ \textless \ \frac{3}{4}$

$\displaystyle -\frac{ \sqrt{3}}{2}\ \textless \ cosx\ \textless \ \frac{ \sqrt{3}}{2}$

1)
$\displaystyle cosx\ \textgreater \ - \frac{ \sqrt{3}}{2} 
$

$\displaystyle - \frac{5 \pi }{6}+2 \pi n\ \textless \ x\ \textless \ \frac{5 \pi }{6}+2 \pi n; n\in Z$

2)
$\displaystyle cosx\ \textless \ \frac{ \sqrt{3}}{2}$

$\displaystyle \frac{ \pi }{6}+2 \pi n\ \textless \ x\ \textless \ \frac{11 \pi }{6}+2 \pi n; n\in Z$

объединим промежутки

$\displaystyle (- \frac{5 \pi }{6}+2 \pi n; - \frac{\pi }{6}+2 \pi n) ( \frac{ \pi }{6}+2 \pi n; \frac{5 \pi }{6}+2 \pi n); n\in Z$

10 votes Thanks 15

oganesbagoyan

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Task/1196740
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3-4 cos^2x>0 как решить неравенство ?
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3-4cos²x > 0 ;
3- 4*(1+cos2x)/2  > 0 ;
3- 2-2cos2x  > 0 ;
cos2x < 1/2 ;
2πn + π/3 < 2x < 2π - π/3 + 2πn , n ∈ Z .
πn + π/6 < x < 5π/6 + πn , n ∈ Z .

ответ : ∪ ( πn + π/6 ; 5π/6 + πn ) .
   n ∈ Z .

4 votes Thanks 8

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