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Beliebt Trigonometrie >

cos^2(x)=((4k-5))/(10)

Lösung

cos^{2} (x) = \frac{( 4 k - 5 )}{10}

Lösung

x = arccos (\frac{4 k - 5}{10}) + 2 πn, x = - arccos (\frac{4 k - 5}{10}) + 2 πn, x = arccos (- \frac{4 k - 5}{10}) + 2 πn, x = - arccos (- \frac{4 k - 5}{10}) + 2 πn

Schritte zur Lösung

cos^{2} (x) = \frac{( 4 k - 5 )}{10}

Löse mit Substitution

cos^{2} (x) = \frac{4 k - 5}{10}

Angenommen:

cos (x) = u

u^{2} = \frac{4 k - 5}{10}

u^{2} = \frac{4 k - 5}{10} : u = \frac{4 k - 5}{10}, u = - \frac{4 k - 5}{10}

u^{2} = \frac{4 k - 5}{10}

Für

x^{2} = f (a)

sind die Lösungen

x = f (a), - f (a)

u = \frac{4 k - 5}{10}, u = - \frac{4 k - 5}{10}

Setze in

u = cos (x)

ein

cos (x) = \frac{4 k - 5}{10}, cos (x) = - \frac{4 k - 5}{10}

cos (x) = \frac{4 k - 5}{10}, cos (x) = - \frac{4 k - 5}{10}

cos (x) = \frac{4 k - 5}{10} : x = arccos (\frac{4 k - 5}{10}) + 2 πn, x = - arccos (\frac{4 k - 5}{10}) + 2 πn

cos (x) = \frac{4 k - 5}{10}

Wende die Eigenschaften der Trigonometrie an

cos (x) = \frac{4 k - 5}{10}

Allgemeine Lösung für

cos (x) = \frac{4 k - 5}{10}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos (\frac{4 k - 5}{10}) + 2 πn, x = - arccos (\frac{4 k - 5}{10}) + 2 πn

x = arccos (\frac{4 k - 5}{10}) + 2 πn, x = - arccos (\frac{4 k - 5}{10}) + 2 πn

cos (x) = - \frac{4 k - 5}{10} : x = arccos (- \frac{4 k - 5}{10}) + 2 πn, x = - arccos (- \frac{4 k - 5}{10}) + 2 πn

cos (x) = - \frac{4 k - 5}{10}

Wende die Eigenschaften der Trigonometrie an

cos (x) = - \frac{4 k - 5}{10}

Allgemeine Lösung für

cos (x) = - \frac{4 k - 5}{10}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos (- \frac{4 k - 5}{10}) + 2 πn, x = - arccos (- \frac{4 k - 5}{10}) + 2 πn

x = arccos (- \frac{4 k - 5}{10}) + 2 πn, x = - arccos (- \frac{4 k - 5}{10}) + 2 πn

Kombiniere alle Lösungen

x = arccos (\frac{4 k - 5}{10}) + 2 πn, x = - arccos (\frac{4 k - 5}{10}) + 2 πn, x = arccos (- \frac{4 k - 5}{10}) + 2 πn, x = - arccos (- \frac{4 k - 5}{10}) + 2 πn

Graph

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