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Populaire Trigonométrie >

tan^2(x)sec(x)-1=0

Solution

tan^{2} (x) sec (x) - 1 = 0

Solution

x = 0.71532 \dots + 2 πn, x = 2 π - 0.71532 \dots + 2 πn

Degrés

x = 40.98531 \dots^{\circ} + 36 0^{\circ} n, x = 319.01468 \dots^{\circ} + 36 0^{\circ} n

étapes des solutions

tan^{2} (x) sec (x) - 1 = 0

Récrire en utilisant des identités trigonométriques

- 1 + sec (x) tan^{2} (x)

Utiliser l'identité hyperbolique:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

= - 1 + sec (x) (sec^{2} (x) - 1)

- 1 + (- 1 + sec^{2} (x)) sec (x) = 0

Résoudre par substitution

- 1 + (- 1 + sec^{2} (x)) sec (x) = 0

Soit :

sec (x) = u

- 1 + (- 1 + u^{2}) u = 0

- 1 + (- 1 + u^{2}) u = 0 : u \approx 1.32471 \dots

- 1 + (- 1 + u^{2}) u = 0

Développer

- 1 + (- 1 + u^{2}) u : - 1 - u + u^{3}

- 1 + (- 1 + u^{2}) u

= - 1 + u (- 1 + u^{2})

Développer

u (- 1 + u^{2}) : - u + u^{3}

u (- 1 + u^{2})

Appliquer la loi de la distribution:

a (b + c) = ab + a c

a = u, b = - 1, c = u^{2}

= u (- 1) + u u^{2}

Appliquer les règles des moins et des plus

+ (- a) = - a

= - 1 \cdot u + u^{2} u

Simplifier

- 1 \cdot u + u^{2} u : - u + u^{3}

- 1 \cdot u + u^{2} u

1 \cdot u = u

1 \cdot u

Multiplier:

1 \cdot u = u

= u

u^{2} u = u^{3}

u^{2} u

Appliquer la règle de l'exposant:

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

u^{2} u = u^{2 + 1}

= u^{2 + 1}

Additionner les nombres :

2 + 1 = 3

= u^{3}

= - u + u^{3}

= - u + u^{3}

= - 1 - u + u^{3}

- 1 - u + u^{3} = 0

Ecrire sous la forme standard

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{3} - u - 1 = 0

Trouver une solution pour

u^{3} - u - 1 = 0

par la méthode de Newton-Raphson:

u \approx 1.32471 \dots

u^{3} - u - 1 = 0

Définition de l'approximation de Newton-Raphson

f (u) = u^{3} - u - 1

Trouver

f^{^{'}} (u) : 3 u^{2} - 1

\frac{d}{d u} (u^{3} - u - 1)

Appliquer la règle de l'addition/soustraction:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{3}) - \frac{d u}{d u} - \frac{d}{d u} (1)

\frac{d}{d u} (u^{3}) = 3 u^{2}

\frac{d}{d u} (u^{3})

Appliquer la règle de la puissance:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 u^{3 - 1}

Simplifier

= 3 u^{2}

\frac{d u}{d u} = 1

\frac{d u}{d u}

Appliquer la dérivée commune:

\frac{d u}{d u} = 1

= 1

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

Dérivée d'une constante:

\frac{d}{d x} (a) = 0

= 0

= 3 u^{2} - 1 - 0

Simplifier

= 3 u^{2} - 1

Soit

u_{0} = 1

Calculer

u_{n + 1}

jusqu'à

Δ u_{n + 1} < 0.000001

u_{1} = 1.5 : Δ u_{1} = 0.5

f (u_{0}) = 1^{3} - 1 - 1 = - 1

f^{^{'}} (u_{0}) = 3 \cdot 1^{2} - 1 = 2

u_{1} = 1.5

Δ u_{1} = ∣ 1.5 - 1 ∣ = 0.5

Δ u_{1} = 0.5

u_{2} = 1.34782 \dots : Δ u_{2} = 0.15217 \dots

f (u_{1}) = 1. 5^{3} - 1.5 - 1 = 0.875

f^{^{'}} (u_{1}) = 3 \cdot 1. 5^{2} - 1 = 5.75

u_{2} = 1.34782 \dots

Δ u_{2} = ∣ 1.34782 \dots - 1.5 ∣ = 0.15217 \dots

Δ u_{2} = 0.15217 \dots

u_{3} = 1.32520 \dots : Δ u_{3} = 0.02262 \dots

f (u_{2}) = 1.34782 \dots^{3} - 1.34782 \dots - 1 = 0.10068 \dots

f^{^{'}} (u_{2}) = 3 \cdot 1.34782 \dots^{2} - 1 = 4.44990 \dots

u_{3} = 1.32520 \dots

Δ u_{3} = ∣ 1.32520 \dots - 1.34782 \dots ∣ = 0.02262 \dots

Δ u_{3} = 0.02262 \dots

u_{4} = 1.32471 \dots : Δ u_{4} = 0.00048 \dots

f (u_{3}) = 1.32520 \dots^{3} - 1.32520 \dots - 1 = 0.00205 \dots

f^{^{'}} (u_{3}) = 3 \cdot 1.32520 \dots^{2} - 1 = 4.26846 \dots

u_{4} = 1.32471 \dots

Δ u_{4} = ∣ 1.32471 \dots - 1.32520 \dots ∣ = 0.00048 \dots

Δ u_{4} = 0.00048 \dots

u_{5} = 1.32471 \dots : Δ u_{5} = 2.16754 E - 7

f (u_{4}) = 1.32471 \dots^{3} - 1.32471 \dots - 1 = 9.24378 E - 7

f^{^{'}} (u_{4}) = 3 \cdot 1.32471 \dots^{2} - 1 = 4.26463 \dots

u_{5} = 1.32471 \dots

Δ u_{5} = ∣ 1.32471 \dots - 1.32471 \dots ∣ = 2.16754 E - 7

Δ u_{5} = 2.16754 E - 7

u \approx 1.32471 \dots

Appliquer une division longue:

\frac{u ^{3} - u - 1}{u - 1.32471 \dots} = u^{2} + 1.32471 \dots u + 0.75487 \dots

u^{2} + 1.32471 \dots u + 0.75487 \dots \approx 0

Trouver une solution pour

u^{2} + 1.32471 \dots u + 0.75487 \dots = 0

par la méthode de Newton-Raphson:

Aucune solution pour

u \in R

u^{2} + 1.32471 \dots u + 0.75487 \dots = 0

Définition de l'approximation de Newton-Raphson

f (u) = u^{2} + 1.32471 \dots u + 0.75487 \dots

Trouver

f^{^{'}} (u) : 2 u + 1.32471 \dots

\frac{d}{d u} (u^{2} + 1.32471 \dots u + 0.75487 \dots)

Appliquer la règle de l'addition/soustraction:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{2}) + \frac{d}{d u} (1.32471 \dots u) + \frac{d}{d u} (0.75487 \dots)

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

Appliquer la règle de la puissance:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

Simplifier

= 2 u

\frac{d}{d u} (1.32471 \dots u) = 1.32471 \dots

\frac{d}{d u} (1.32471 \dots u)

Retirer la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 1.32471 \dots \frac{d u}{d u}

Appliquer la dérivée commune:

\frac{d u}{d u} = 1

= 1.32471 \dots \cdot 1

Simplifier

= 1.32471 \dots

\frac{d}{d u} (0.75487 \dots) = 0

\frac{d}{d u} (0.75487 \dots)

Dérivée d'une constante:

\frac{d}{d x} (a) = 0

= 0

= 2 u + 1.32471 \dots + 0

Simplifier

= 2 u + 1.32471 \dots

Soit

u_{0} = - 1

Calculer

u_{n + 1}

jusqu'à

Δ u_{n + 1} < 0.000001

u_{1} = - 0.36299 \dots : Δ u_{1} = 0.63700 \dots

f (u_{0}) = (- 1)^{2} + 1.32471 \dots (- 1) + 0.75487 \dots = 0.43015 \dots

f^{^{'}} (u_{0}) = 2 (- 1) + 1.32471 \dots = - 0.67528 \dots

u_{1} = - 0.36299 \dots

Δ u_{1} = ∣ - 0.36299 \dots - (- 1) ∣ = 0.63700 \dots

Δ u_{1} = 0.63700 \dots

u_{2} = - 1.04072 \dots : Δ u_{2} = 0.67772 \dots

f (u_{1}) = (- 0.36299 \dots)^{2} + 1.32471 \dots (- 0.36299 \dots) + 0.75487 \dots = 0.40577 \dots

f^{^{'}} (u_{1}) = 2 (- 0.36299 \dots) + 1.32471 \dots = 0.59873 \dots

u_{2} = - 1.04072 \dots

Δ u_{2} = ∣ - 1.04072 \dots - (- 0.36299 \dots) ∣ = 0.67772 \dots

Δ u_{2} = 0.67772 \dots

u_{3} = - 0.43374 \dots : Δ u_{3} = 0.60697 \dots

f (u_{2}) = (- 1.04072 \dots)^{2} + 1.32471 \dots (- 1.04072 \dots) + 0.75487 \dots = 0.45931 \dots

f^{^{'}} (u_{2}) = 2 (- 1.04072 \dots) + 1.32471 \dots = - 0.75672 \dots

u_{3} = - 0.43374 \dots

Δ u_{3} = ∣ - 0.43374 \dots - (- 1.04072 \dots) ∣ = 0.60697 \dots

Δ u_{3} = 0.60697 \dots

u_{4} = - 1.23950 \dots : Δ u_{4} = 0.80576 \dots

f (u_{3}) = (- 0.43374 \dots)^{2} + 1.32471 \dots (- 0.43374 \dots) + 0.75487 \dots = 0.36842 \dots

f^{^{'}} (u_{3}) = 2 (- 0.43374 \dots) + 1.32471 \dots = 0.45723 \dots

u_{4} = - 1.23950 \dots

Δ u_{4} = ∣ - 1.23950 \dots - (- 0.43374 \dots) ∣ = 0.80576 \dots

Δ u_{4} = 0.80576 \dots

u_{5} = - 0.67703 \dots : Δ u_{5} = 0.56247 \dots

f (u_{4}) = (- 1.23950 \dots)^{2} + 1.32471 \dots (- 1.23950 \dots) + 0.75487 \dots = 0.64925 \dots

f^{^{'}} (u_{4}) = 2 (- 1.23950 \dots) + 1.32471 \dots = - 1.15429 \dots

u_{5} = - 0.67703 \dots

Δ u_{5} = ∣ - 0.67703 \dots - (- 1.23950 \dots) ∣ = 0.56247 \dots

Δ u_{5} = 0.56247 \dots

u_{6} = 10.09982 \dots : Δ u_{6} = 10.77686 \dots

f (u_{5}) = (- 0.67703 \dots)^{2} + 1.32471 \dots (- 0.67703 \dots) + 0.75487 \dots = 0.31637 \dots

f^{^{'}} (u_{5}) = 2 (- 0.67703 \dots) + 1.32471 \dots = - 0.02935 \dots

u_{6} = 10.09982 \dots

Δ u_{6} = ∣ 10.09982 \dots - (- 0.67703 \dots) ∣ = 10.77686 \dots

Δ u_{6} = 10.77686 \dots

u_{7} = 4.70404 \dots : Δ u_{7} = 5.39578 \dots

f (u_{6}) = 10.09982 \dots^{2} + 1.32471 \dots \cdot 10.09982 \dots + 0.75487 \dots = 116.14080 \dots

f^{^{'}} (u_{6}) = 2 \cdot 10.09982 \dots + 1.32471 \dots = 21.52437 \dots

u_{7} = 4.70404 \dots

Δ u_{7} = ∣ 4.70404 \dots - 10.09982 \dots ∣ = 5.39578 \dots

Δ u_{7} = 5.39578 \dots

u_{8} = 1.99138 \dots : Δ u_{8} = 2.71265 \dots

f (u_{7}) = 4.70404 \dots^{2} + 1.32471 \dots \cdot 4.70404 \dots + 0.75487 \dots = 29.11445 \dots

f^{^{'}} (u_{7}) = 2 \cdot 4.70404 \dots + 1.32471 \dots = 10.73280 \dots

u_{8} = 1.99138 \dots

Δ u_{8} = ∣ 1.99138 \dots - 4.70404 \dots ∣ = 2.71265 \dots

Δ u_{8} = 2.71265 \dots

u_{9} = 0.60494 \dots : Δ u_{9} = 1.38644 \dots

f (u_{8}) = 1.99138 \dots^{2} + 1.32471 \dots \cdot 1.99138 \dots + 0.75487 \dots = 7.35852 \dots

f^{^{'}} (u_{8}) = 2 \cdot 1.99138 \dots + 1.32471 \dots = 5.30749 \dots

u_{9} = 0.60494 \dots

Δ u_{9} = ∣ 0.60494 \dots - 1.99138 \dots ∣ = 1.38644 \dots

Δ u_{9} = 1.38644 \dots

u_{10} = - 0.15344 \dots : Δ u_{10} = 0.75838 \dots

f (u_{9}) = 0.60494 \dots^{2} + 1.32471 \dots \cdot 0.60494 \dots + 0.75487 \dots = 1.92221 \dots

f^{^{'}} (u_{9}) = 2 \cdot 0.60494 \dots + 1.32471 \dots = 2.53460 \dots

u_{10} = - 0.15344 \dots

Δ u_{10} = ∣ - 0.15344 \dots - 0.60494 \dots ∣ = 0.75838 \dots

Δ u_{10} = 0.75838 \dots

u_{11} = - 0.71852 \dots : Δ u_{11} = 0.56507 \dots

f (u_{10}) = (- 0.15344 \dots)^{2} + 1.32471 \dots (- 0.15344 \dots) + 0.75487 \dots = 0.57515 \dots

f^{^{'}} (u_{10}) = 2 (- 0.15344 \dots) + 1.32471 \dots = 1.01783 \dots

u_{11} = - 0.71852 \dots

Δ u_{11} = ∣ - 0.71852 \dots - (- 0.15344 \dots) ∣ = 0.56507 \dots

Δ u_{11} = 0.56507 \dots

u_{12} = 2.12427 \dots : Δ u_{12} = 2.84279 \dots

f (u_{11}) = (- 0.71852 \dots)^{2} + 1.32471 \dots (- 0.71852 \dots) + 0.75487 \dots = 0.31931 \dots

f^{^{'}} (u_{11}) = 2 (- 0.71852 \dots) + 1.32471 \dots = - 0.11232 \dots

u_{12} = 2.12427 \dots

Δ u_{12} = ∣ 2.12427 \dots - (- 0.71852 \dots) ∣ = 2.84279 \dots

Δ u_{12} = 2.84279 \dots

Impossible de trouver une solution

La solution est

u \approx 1.32471 \dots

Remplacer

u = sec (x)

sec (x) \approx 1.32471 \dots

sec (x) \approx 1.32471 \dots

sec (x) = 1.32471 \dots : x = arcsec (1.32471 \dots) + 2 πn, x = 2 π - arcsec (1.32471 \dots) + 2 πn

sec (x) = 1.32471 \dots

Appliquer les propriétés trigonométriques inverses

sec (x) = 1.32471 \dots

Solutions générales pour

sec (x) = 1.32471 \dots

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = 2 π - arcsec (a) + 2 πn

x = arcsec (1.32471 \dots) + 2 πn, x = 2 π - arcsec (1.32471 \dots) + 2 πn

x = arcsec (1.32471 \dots) + 2 πn, x = 2 π - arcsec (1.32471 \dots) + 2 πn

Combiner toutes les solutions

x = arcsec (1.32471 \dots) + 2 πn, x = 2 π - arcsec (1.32471 \dots) + 2 πn

Montrer les solutions sous la forme décimale

x = 0.71532 \dots + 2 πn, x = 2 π - 0.71532 \dots + 2 πn

Graphe

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