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Beliebt Trigonometrie >

0<= x<= 2piarccos(1/2)

Lösung

0 \leq x \leq 2 π arccos (\frac{1}{2})

Lösung

0 \leq x \leq \frac{2 π ^{2}}{3}

+2

Intervall-Notation

[0, \frac{2 π ^{2}}{3}]

Dezimale

0 \leq x \leq 6.57973 \dots

Schritte zur Lösung

0 \leq x \leq 2 π arccos (\frac{1}{2})

2 π arccos (\frac{1}{2}) = \frac{2 π ^{2}}{3}

2 π arccos (\frac{1}{2})

Vereinfache

arccos (\frac{1}{2}) : \frac{π}{3}

arccos (\frac{1}{2})

Verwende die folgende triviale Identität:

arccos (\frac{1}{2}) = \frac{π}{3}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{3}

= 2 π \frac{π}{3}

Vereinfache

2 π \frac{π}{3}

Multipliziere Brüche:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2 π}{3}

π 2 π = 2 π^{2}

π 2 π

Wende Exponentenregel an:

a^{b} \cdot a^{c} = a^{b + c}

ππ = π^{1 + 1}

= 2 π^{1 + 1}

Addiere die Zahlen:

1 + 1 = 2

= 2 π^{2}

= \frac{2 π ^{2}}{3}

= \frac{2 π ^{2}}{3}

0 \leq x \leq \frac{2 π ^{2}}{3}

Graph

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