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√sin2(−1)≤1−sin(x)
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Solution
−π−arcsin(−sin(1)+1)+2πn≤x≤arcsin(−sin(1)+1)+2πn
+2
Interval Notation
[−π−arcsin(−sin(1)+1)+2πn,arcsin(−sin(1)+1)+2πn]
Decimal
−3.30079…+2πn≤x≤0.15920…+2πn
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√sin2(−1)≤1−sin(x)
Tausche die Seiten
1−sin(x)≥√sin2(−1)
Vereinfache √sin2(−1):sin(1)
1−sin(x)≥sin(1)
Verschiebe 1auf die rechte Seite
−sin(x)≥sin(1)−1
Multipliziere beide Seiten mit −1
sin(x)≤−sin(1)+1
Für sin(x)≤a, wenn −1<a<1 dann −π−arcsin(a)+2πn≤x≤arcsin(a)+2πn