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شائع علم المثلثات >

sin^2(x)=sec(x)

الحلّ

sin^{2} (x) = sec (x)

الحلّ

x \in R لا يوجد حلّ لـ

خطوات الحلّ

sin^{2} (x) = sec (x)

ربّع الطرفين

(sin^{2} (x))^{2} = sec^{2} (x)

من الطرفين

sec^{2} (x)

اطرح

sin^{4} (x) - sec^{2} (x) = 0

sin, cos :

عبّر بواسطة

- sec^{2} (x) + sin^{4} (x)

sec (x) = \frac{1}{cos ( x )}

:Use the basic trigonometric identity

= - (\frac{1}{cos ( x )})^{2} + sin^{4} (x)

- (\frac{1}{cos ( x )})^{2} + sin^{4} (x)

بسّط:

\frac{- 1 + sin ^{4} ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

- (\frac{1}{cos ( x )})^{2} + sin^{4} (x)

(\frac{1}{cos ( x )})^{2} = \frac{1}{cos ^{2} ( x )}

(\frac{1}{cos ( x )})^{2}

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

:فعّل قانون القوى

= \frac{1 ^{2}}{cos ^{2} ( x )}

1^{a} = 1

فعّل القانون

1^{2} = 1

= \frac{1}{cos ^{2} ( x )}

= - \frac{1}{cos ^{2} ( x )} + sin^{4} (x)

sin^{4} (x) = \frac{sin ^{4} ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

:حوّل الأعداد لكسور

= - \frac{1}{cos ^{2} ( x )} + \frac{sin ^{4} ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

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

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{- 1 + sin ^{4} ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

= \frac{- 1 + sin ^{4} ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

\frac{- 1 + cos ^{2} ( x ) sin ^{4} ( x )}{cos ^{2} ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- 1 + cos^{2} (x) sin^{4} (x) = 0

- 1 + cos^{2} (x) sin^{4} (x)

حلل إلى عوامل:

(sin^{2} (x) cos (x) + 1) (sin^{2} (x) cos (x) - 1)

- 1 + cos^{2} (x) sin^{4} (x)

- 1 + (sin^{2} (x) cos (x))^{2}

كـ

- 1 + sin^{4} (x) cos^{2} (x)

اكتب مجددًا

- 1 + sin^{4} (x) cos^{2} (x)

a^{b c} = (a^{b})^{c}

:فعّل قانون القوى

sin^{4} (x) = (sin^{2} (x))^{2}

= - 1 + (sin^{2} (x))^{2} cos^{2} (x)

a^{m} b^{m} = (ab)^{m}

:فعّل قانون القوى

(sin^{2} (x))^{2} cos^{2} (x) = (sin^{2} (x) cos (x))^{2}

= - 1 + (sin^{2} (x) cos (x))^{2}

= - 1 + (sin^{2} (x) cos (x))^{2}

x^{2} - y^{2} = (x + y) (x - y)

فعّل قانون فرق المربّعات

- 1 + (sin^{2} (x) cos (x))^{2} = (sin^{2} (x) cos (x) + 1) (sin^{2} (x) cos (x) - 1)

= (sin^{2} (x) cos (x) + 1) (sin^{2} (x) cos (x) - 1)

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

حلّ كل جزء على حدة

sin^{2} (x) cos (x) + 1 = 0 or sin^{2} (x) cos (x) - 1 = 0

sin^{2} (x) cos (x) + 1 = 0 :

لا يوجد حلّ

sin^{2} (x) cos (x) + 1 = 0

Rewrite using trig identities

1 + cos (x) sin^{2} (x)

cos^{2} (x) + sin^{2} (x) = 1

:فعّل نطريّة فيتاغوروس

sin^{2} (x) = 1 - cos^{2} (x)

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

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

بالاستعانة بطريقة التعويض

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

cos (x) = u :

على افتراض أنّ

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

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

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

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

وسّع:

1 + u - u^{3}

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

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

u (1 - u^{2})

وسٌع:

u - u^{3}

u (1 - u^{2})

a (b - c) = ab - a c

: افتح أقواس بالاستعانة بـ

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

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

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

1 \cdot u - u^{2} u

بسّط:

u - u^{3}

1 \cdot u - u^{2} u

1 \cdot u = u

1 \cdot u

1 \cdot u = u :

اضرب

= u

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

u^{2} u

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

:فعّل قانون القوى

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

= u^{2 + 1}

2 + 1 = 3 :

اجمع الأعداد

= u^{3}

= u - u^{3}

= u - u^{3}

= 1 + u - u^{3}

1 + u - u^{3} = 0

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

اكتب بالصورة الاعتياديّة

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

بطريقة نيوتون ريبسون

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

جدّ حلًا لـ:

u \approx 1.32471 \dots

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

تعريف تقريب نيوتن-ريبسون

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

f^{^{'}} (u)

جد:

- 3 u^{2} + 1

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

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

:استعمل قانون الجمع

= - \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})

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

:استعمل قانون الأسس

= 3 u^{3 - 1}

بسّط

= 3 u^{2}

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

\frac{d u}{d u}

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

:استعمل المشتقة الأساسية

= 1

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

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

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

:مشتقة الثابت

= 0

= - 3 u^{2} + 1 + 0

بسّط

= - 3 u^{2} + 1

u_{0} = 1

استبدل

Δ u_{n + 1} < 0.000001

حتّى

u_{n + 1}

احسب

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

فعّل القسمة الطويلة:

\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

بطريقة نيوتون ريبسون

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

جدّ حلًا لـ:

u \in R

لا يوجد حلّ لـ

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

تعريف تقريب نيوتن-ريبسون

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

f^{^{'}} (u)

جد:

- 2 u - 1.32471 \dots

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

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

:استعمل قانون الجمع

= - \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})

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

:استعمل قانون الأسس

= 2 u^{2 - 1}

بسّط

= 2 u

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

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

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

:استخرج الثابت

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

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

:استعمل المشتقة الأساسية

= 1.32471 \dots \cdot 1

بسّط

= 1.32471 \dots

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

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

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

:مشتقة الثابت

= 0

= - 2 u - 1.32471 \dots - 0

بسّط

= - 2 u - 1.32471 \dots

u_{0} = - 1

استبدل

Δ u_{n + 1} < 0.000001

حتّى

u_{n + 1}

احسب

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

لا يمكن إيجاد حلّ

الحل للمعادلة هو

u \approx 1.32471 \dots

u = cos (x)

استبدل مجددًا

cos (x) \approx 1.32471 \dots

cos (x) \approx 1.32471 \dots

cos (x) = 1.32471 \dots :

لا يوجد حلّ

cos (x) = 1.32471 \dots

- 1 \leq cos (x) \leq 1

لا يوجد حلّ

وحّد الحلول

لا يوجد حلّ

sin^{2} (x) cos (x) - 1 = 0 :

لا يوجد حلّ

sin^{2} (x) cos (x) - 1 = 0

Rewrite using trig identities

- 1 + cos (x) sin^{2} (x)

cos^{2} (x) + sin^{2} (x) = 1

:فعّل نطريّة فيتاغوروس

sin^{2} (x) = 1 - cos^{2} (x)

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

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

بالاستعانة بطريقة التعويض

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

cos (x) = u :

على افتراض أنّ

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

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

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

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

وسّع:

- 1 + u - u^{3}

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

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

u (1 - u^{2})

وسٌع:

u - u^{3}

u (1 - u^{2})

a (b - c) = ab - a c

: افتح أقواس بالاستعانة بـ

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

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

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

1 \cdot u - u^{2} u

بسّط:

u - u^{3}

1 \cdot u - u^{2} u

1 \cdot u = u

1 \cdot u

1 \cdot u = u :

اضرب

= u

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

u^{2} u

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

:فعّل قانون القوى

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

= u^{2 + 1}

2 + 1 = 3 :

اجمع الأعداد

= u^{3}

= u - u^{3}

= u - u^{3}

= - 1 + u - u^{3}

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

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

اكتب بالصورة الاعتياديّة

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

بطريقة نيوتون ريبسون

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

جدّ حلًا لـ:

u \approx - 1.32471 \dots

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

تعريف تقريب نيوتن-ريبسون

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

f^{^{'}} (u)

جد:

- 3 u^{2} + 1

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

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

:استعمل قانون الجمع

= - \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})

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

:استعمل قانون الأسس

= 3 u^{3 - 1}

بسّط

= 3 u^{2}

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

\frac{d u}{d u}

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

:استعمل المشتقة الأساسية

= 1

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

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

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

:مشتقة الثابت

= 0

= - 3 u^{2} + 1 - 0

بسّط

= - 3 u^{2} + 1

u_{0} = - 1

استبدل

Δ u_{n + 1} < 0.000001

حتّى

u_{n + 1}

احسب

u_{1} = - 1.5 : Δ u_{1} = 0.5

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

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

فعّل القسمة الطويلة:

\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

بطريقة نيوتون ريبسون

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

جدّ حلًا لـ:

u \in R

لا يوجد حلّ لـ

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

تعريف تقريب نيوتن-ريبسون

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

f^{^{'}} (u)

جد:

- 2 u + 1.32471 \dots

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

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

:استعمل قانون الجمع

= - \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})

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

:استعمل قانون الأسس

= 2 u^{2 - 1}

بسّط

= 2 u

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

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

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

:استخرج الثابت

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

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

:استعمل المشتقة الأساسية

= 1.32471 \dots \cdot 1

بسّط

= 1.32471 \dots

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

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

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

:مشتقة الثابت

= 0

= - 2 u + 1.32471 \dots - 0

بسّط

= - 2 u + 1.32471 \dots

u_{0} = 1

استبدل

Δ u_{n + 1} < 0.000001

حتّى

u_{n + 1}

احسب

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

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

f^{^{'}} (u_{0}) = - 2 \cdot 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 \cdot 0.36299 \dots - 0.75487 \dots = - 0.40577 \dots

f^{^{'}} (u_{1}) = - 2 \cdot 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 \cdot 1.04072 \dots - 0.75487 \dots = - 0.45931 \dots

f^{^{'}} (u_{2}) = - 2 \cdot 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 \cdot 0.43374 \dots - 0.75487 \dots = - 0.36842 \dots

f^{^{'}} (u_{3}) = - 2 \cdot 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 \cdot 1.23950 \dots - 0.75487 \dots = - 0.64925 \dots

f^{^{'}} (u_{4}) = - 2 \cdot 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 \cdot 0.67703 \dots - 0.75487 \dots = - 0.31637 \dots

f^{^{'}} (u_{5}) = - 2 \cdot 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 (- 10.09982 \dots) - 0.75487 \dots = - 116.14080 \dots

f^{^{'}} (u_{6}) = - 2 (- 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 (- 4.70404 \dots) - 0.75487 \dots = - 29.11445 \dots

f^{^{'}} (u_{7}) = - 2 (- 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 (- 1.99138 \dots) - 0.75487 \dots = - 7.35852 \dots

f^{^{'}} (u_{8}) = - 2 (- 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 (- 0.60494 \dots) - 0.75487 \dots = - 1.92221 \dots

f^{^{'}} (u_{9}) = - 2 (- 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 \cdot 0.15344 \dots - 0.75487 \dots = - 0.57515 \dots

f^{^{'}} (u_{10}) = - 2 \cdot 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 \cdot 0.71852 \dots - 0.75487 \dots = - 0.31931 \dots

f^{^{'}} (u_{11}) = - 2 \cdot 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

لا يمكن إيجاد حلّ

الحل للمعادلة هو

u \approx - 1.32471 \dots

u = cos (x)

استبدل مجددًا

cos (x) \approx - 1.32471 \dots

cos (x) \approx - 1.32471 \dots

cos (x) = - 1.32471 \dots :

لا يوجد حلّ

cos (x) = - 1.32471 \dots

- 1 \leq cos (x) \leq 1

لا يوجد حلّ

وحّد الحلول

لا يوجد حلّ

وحّد الحلول

لا يوجد حلّ

تأكّد من صحّة الحلول عن طريق تعويضها في المعادلة الأصليّة

للتحقّق من دقّة الحلول

sin^{2} (x) = sec (x)

عوّض الحلول في

إلغي الحلول التي تعطي قضيّة كذب

x \in R لا يوجد حلّ لـ

رسم

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3cos^2(x)-sin^2(x)-sin^2(x)=0

3sin(a)+cos(a)=1

4cos^2(x)+sqrt(3)=2(sqrt(3)+1)

2tan^2(x)+cot^2(x)-3=0