Rozwiąż nierówność dla :
[ 2 cos x - 1]/ [ 2 + cos 3x ] > = 0
Ponieważ
- 1 <= cos 3x < = 1 , zatem
2 + cos 3 x > = 0
oraz
2 cos x - 1 > = 0
2 cos x >= 1 / : 2
cos x > = 1/2
Odp.
x należy do < 0; pi/3 > u < (5 pi)/3 ; 2 pi >
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[ 2 cos x - 1]/ [ 2 + cos 3x ] > = 0
Ponieważ
- 1 <= cos 3x < = 1 , zatem
2 + cos 3 x > = 0
oraz
2 cos x - 1 > = 0
2 cos x >= 1 / : 2
cos x > = 1/2
Odp.
x należy do < 0; pi/3 > u < (5 pi)/3 ; 2 pi >
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