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شائع علم المثلثات >

2csc^2(x)=1+cos(x)

الحلّ

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

الحلّ

x \in R لا يوجد حلّ لـ

خطوات الحلّ

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

من الطرفين

1 + cos (x)

اطرح

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

Rewrite using trig identities

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

csc (x) = \frac{1}{sin ( x )}

:Use the basic trigonometric identity

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

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

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

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

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

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

:فعّل قانون القوى

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

1^{a} = 1

فعّل القانون

1^{2} = 1

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

= 2 \cdot \frac{1}{sin ^{2} ( x )}

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

:اضرب كسور

= \frac{1 \cdot 2}{sin ^{2} ( x )}

1 \cdot 2 = 2 :

اضرب الأعداد

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

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

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

:فعّل نطريّة فيتاغوروس

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

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

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

بالاستعانة بطريقة التعويض

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

cos (x) = u :

على افتراض أنّ

- 1 - u + \frac{2}{1 - u ^{2}} = 0

- 1 - u + \frac{2}{1 - u ^{2}} = 0 : u \approx - 1.83928 \dots

- 1 - u + \frac{2}{1 - u ^{2}} = 0

1 - u^{2}

اضرب الطرفين بـ

- 1 - u + \frac{2}{1 - u ^{2}} = 0

1 - u^{2}

اضرب الطرفين بـ

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

بسّط

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

- 1 \cdot (1 - u^{2})

بسّط:

- (1 - u^{2})

- 1 \cdot (1 - u^{2})

1 \cdot (1 - u^{2}) = (1 - u^{2}) :

اضرب

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

\frac{2}{1 - u ^{2}} (1 - u^{2})

بسّط:

2

\frac{2}{1 - u ^{2}} (1 - u^{2})

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

:اضرب كسور

= \frac{2 ( 1 - u ^{2} )}{1 - u ^{2}}

1 - u^{2} :

إلغ العوامل المشتركة

= 2

0 \cdot (1 - u^{2})

بسّط:

0

0 \cdot (1 - u^{2})

0 \cdot a = 0

فعّل القانون

= 0

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

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

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

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

حلّ:

u \approx - 1.83928 \dots

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

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

وسّع:

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

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

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

- (1 - u^{2})

افتح أقواس

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

فعّل قوانين سالب-موجب

- (- a) = a, - (a) = - a

= - 1 + u^{2}

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

فعّل قوانين سالب-موجب

- (- a) = a

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

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

بسّط:

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

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

جمّع التعابير المتشابهة

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

- 1 + 2 = 1 :

اطرح/اجمع الأعداد

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

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

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

بطريقة نيوتون ريبسون

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

جدّ حلًا لـ:

u \approx - 1.83928 \dots

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

تعريف تقريب نيوتن-ريبسون

f (u) = u^{3} + u^{2} - u + 1

f^{^{'}} (u)

جد:

3 u^{2} + 2 u - 1

\frac{d}{d u} (u^{3} + u^{2} - u + 1)

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

:استعمل قانون الجمع

= \frac{d}{d u} (u^{3}) + \frac{d}{d u} (u^{2}) - \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}{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 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} + 2 u - 1 + 0

بسّط

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

u_{0} = - 4

استبدل

Δ u_{n + 1} < 0.000001

حتّى

u_{n + 1}

احسب

u_{1} = - 2.89743 \dots : Δ u_{1} = 1.10256 \dots

f (u_{0}) = (- 4)^{3} + (- 4)^{2} - (- 4) + 1 = - 43

f^{^{'}} (u_{0}) = 3 (- 4)^{2} + 2 (- 4) - 1 = 39

u_{1} = - 2.89743 \dots

Δ u_{1} = ∣ - 2.89743 \dots - (- 4) ∣ = 1.10256 \dots

Δ u_{1} = 1.10256 \dots

u_{2} = - 2.24319 \dots : Δ u_{2} = 0.65423 \dots

f (u_{1}) = (- 2.89743 \dots)^{3} + (- 2.89743 \dots)^{2} - (- 2.89743 \dots) + 1 = - 12.03179 \dots

f^{^{'}} (u_{1}) = 3 (- 2.89743 \dots)^{2} + 2 (- 2.89743 \dots) - 1 = 18.39053 \dots

u_{2} = - 2.24319 \dots

Δ u_{2} = ∣ - 2.24319 \dots - (- 2.89743 \dots) ∣ = 0.65423 \dots

Δ u_{2} = 0.65423 \dots

u_{3} = - 1.92970 \dots : Δ u_{3} = 0.31349 \dots

f (u_{2}) = (- 2.24319 \dots)^{3} + (- 2.24319 \dots)^{2} - (- 2.24319 \dots) + 1 = - 3.01249 \dots

f^{^{'}} (u_{2}) = 3 (- 2.24319 \dots)^{2} + 2 (- 2.24319 \dots) - 1 = 9.60940 \dots

u_{3} = - 1.92970 \dots

Δ u_{3} = ∣ - 1.92970 \dots - (- 2.24319 \dots) ∣ = 0.31349 \dots

Δ u_{3} = 0.31349 \dots

u_{4} = - 1.84537 \dots : Δ u_{4} = 0.08433 \dots

f (u_{3}) = (- 1.92970 \dots)^{3} + (- 1.92970 \dots)^{2} - (- 1.92970 \dots) + 1 = - 0.53228 \dots

f^{^{'}} (u_{3}) = 3 (- 1.92970 \dots)^{2} + 2 (- 1.92970 \dots) - 1 = 6.31186 \dots

u_{4} = - 1.84537 \dots

Δ u_{4} = ∣ - 1.84537 \dots - (- 1.92970 \dots) ∣ = 0.08433 \dots

Δ u_{4} = 0.08433 \dots

u_{5} = - 1.83931 \dots : Δ u_{5} = 0.00605 \dots

f (u_{4}) = (- 1.84537 \dots)^{3} + (- 1.84537 \dots)^{2} - (- 1.84537 \dots) + 1 = - 0.03345 \dots

f^{^{'}} (u_{4}) = 3 (- 1.84537 \dots)^{2} + 2 (- 1.84537 \dots) - 1 = 5.52545 \dots

u_{5} = - 1.83931 \dots

Δ u_{5} = ∣ - 1.83931 \dots - (- 1.84537 \dots) ∣ = 0.00605 \dots

Δ u_{5} = 0.00605 \dots

u_{6} = - 1.83928 \dots : Δ u_{6} = 0.00003 \dots

f (u_{5}) = (- 1.83931 \dots)^{3} + (- 1.83931 \dots)^{2} - (- 1.83931 \dots) + 1 = - 0.00016 \dots

f^{^{'}} (u_{5}) = 3 (- 1.83931 \dots)^{2} + 2 (- 1.83931 \dots) - 1 = 5.47062 \dots

u_{6} = - 1.83928 \dots

Δ u_{6} = ∣ - 1.83928 \dots - (- 1.83931 \dots) ∣ = 0.00003 \dots

Δ u_{6} = 0.00003 \dots

u_{7} = - 1.83928 \dots : Δ u_{7} = 7.61448 E - 10

f (u_{6}) = (- 1.83928 \dots)^{3} + (- 1.83928 \dots)^{2} - (- 1.83928 \dots) + 1 = - 4.16539 E - 9

f^{^{'}} (u_{6}) = 3 (- 1.83928 \dots)^{2} + 2 (- 1.83928 \dots) - 1 = 5.47035 \dots

u_{7} = - 1.83928 \dots

Δ u_{7} = ∣ - 1.83928 \dots - (- 1.83928 \dots) ∣ = 7.61448 E - 10

Δ u_{7} = 7.61448 E - 10

u \approx - 1.83928 \dots

فعّل القسمة الطويلة:

\frac{u ^{3} + u ^{2} - u + 1}{u + 1.83928 \dots} = u^{2} - 0.83928 \dots u + 0.54368 \dots

u^{2} - 0.83928 \dots u + 0.54368 \dots \approx 0

بطريقة نيوتون ريبسون

u^{2} - 0.83928 \dots u + 0.54368 \dots = 0

جدّ حلًا لـ:

u \in R

لا يوجد حلّ لـ

u^{2} - 0.83928 \dots u + 0.54368 \dots = 0

تعريف تقريب نيوتن-ريبسون

f (u) = u^{2} - 0.83928 \dots u + 0.54368 \dots

f^{^{'}} (u)

جد:

2 u - 0.83928 \dots

\frac{d}{d u} (u^{2} - 0.83928 \dots u + 0.54368 \dots)

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

:استعمل قانون الجمع

= \frac{d}{d u} (u^{2}) - \frac{d}{d u} (0.83928 \dots u) + \frac{d}{d u} (0.54368 \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} (0.83928 \dots u) = 0.83928 \dots

\frac{d}{d u} (0.83928 \dots u)

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

:استخرج الثابت

= 0.83928 \dots \frac{d u}{d u}

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

:استعمل المشتقة الأساسية

= 0.83928 \dots \cdot 1

بسّط

= 0.83928 \dots

\frac{d}{d u} (0.54368 \dots) = 0

\frac{d}{d u} (0.54368 \dots)

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

:مشتقة الثابت

= 0

= 2 u - 0.83928 \dots + 0

بسّط

= 2 u - 0.83928 \dots

u_{0} = 1

استبدل

Δ u_{n + 1} < 0.000001

حتّى

u_{n + 1}

احسب

u_{1} = 0.39312 \dots : Δ u_{1} = 0.60687 \dots

f (u_{0}) = 1^{2} - 0.83928 \dots \cdot 1 + 0.54368 \dots = 0.70440 \dots

f^{^{'}} (u_{0}) = 2 \cdot 1 - 0.83928 \dots = 1.16071 \dots

u_{1} = 0.39312 \dots

Δ u_{1} = ∣ 0.39312 \dots - 1 ∣ = 0.60687 \dots

Δ u_{1} = 0.60687 \dots

u_{2} = 7.33847 \dots : Δ u_{2} = 6.94534 \dots

f (u_{1}) = 0.39312 \dots^{2} - 0.83928 \dots \cdot 0.39312 \dots + 0.54368 \dots = 0.36829 \dots

f^{^{'}} (u_{1}) = 2 \cdot 0.39312 \dots - 0.83928 \dots = - 0.05302 \dots

u_{2} = 7.33847 \dots

Δ u_{2} = ∣ 7.33847 \dots - 0.39312 \dots ∣ = 6.94534 \dots

Δ u_{2} = 6.94534 \dots

u_{3} = 3.85249 \dots : Δ u_{3} = 3.48597 \dots

f (u_{2}) = 7.33847 \dots^{2} - 0.83928 \dots \cdot 7.33847 \dots + 0.54368 \dots = 48.23776 \dots

f^{^{'}} (u_{2}) = 2 \cdot 7.33847 \dots - 0.83928 \dots = 13.83765 \dots

u_{3} = 3.85249 \dots

Δ u_{3} = ∣ 3.85249 \dots - 7.33847 \dots ∣ = 3.48597 \dots

Δ u_{3} = 3.48597 \dots

u_{4} = 2.08252 \dots : Δ u_{4} = 1.76996 \dots

f (u_{3}) = 3.85249 \dots^{2} - 0.83928 \dots \cdot 3.85249 \dots + 0.54368 \dots = 12.15204 \dots

f^{^{'}} (u_{3}) = 2 \cdot 3.85249 \dots - 0.83928 \dots = 6.86569 \dots

u_{4} = 2.08252 \dots

Δ u_{4} = ∣ 2.08252 \dots - 3.85249 \dots ∣ = 1.76996 \dots

Δ u_{4} = 1.76996 \dots

u_{5} = 1.14055 \dots : Δ u_{5} = 0.94196 \dots

f (u_{4}) = 2.08252 \dots^{2} - 0.83928 \dots \cdot 2.08252 \dots + 0.54368 \dots = 3.13277 \dots

f^{^{'}} (u_{4}) = 2 \cdot 2.08252 \dots - 0.83928 \dots = 3.32576 \dots

u_{5} = 1.14055 \dots

Δ u_{5} = ∣ 1.14055 \dots - 2.08252 \dots ∣ = 0.94196 \dots

Δ u_{5} = 0.94196 \dots

u_{6} = 0.52515 \dots : Δ u_{6} = 0.61540 \dots

f (u_{5}) = 1.14055 \dots^{2} - 0.83928 \dots \cdot 1.14055 \dots + 0.54368 \dots = 0.88730 \dots

f^{^{'}} (u_{5}) = 2 \cdot 1.14055 \dots - 0.83928 \dots = 1.44183 \dots

u_{6} = 0.52515 \dots

Δ u_{6} = ∣ 0.52515 \dots - 1.14055 \dots ∣ = 0.61540 \dots

Δ u_{6} = 0.61540 \dots

u_{7} = - 1.26953 \dots : Δ u_{7} = 1.79468 \dots

f (u_{6}) = 0.52515 \dots^{2} - 0.83928 \dots \cdot 0.52515 \dots + 0.54368 \dots = 0.37872 \dots

f^{^{'}} (u_{6}) = 2 \cdot 0.52515 \dots - 0.83928 \dots = 0.21102 \dots

u_{7} = - 1.26953 \dots

Δ u_{7} = ∣ - 1.26953 \dots - 0.52515 \dots ∣ = 1.79468 \dots

Δ u_{7} = 1.79468 \dots

u_{8} = - 0.31613 \dots : Δ u_{8} = 0.95339 \dots

f (u_{7}) = (- 1.26953 \dots)^{2} - 0.83928 \dots (- 1.26953 \dots) + 0.54368 \dots = 3.22089 \dots

f^{^{'}} (u_{7}) = 2 (- 1.26953 \dots) - 0.83928 \dots = - 3.37834 \dots

u_{8} = - 0.31613 \dots

Δ u_{8} = ∣ - 0.31613 \dots - (- 1.26953 \dots) ∣ = 0.95339 \dots

Δ u_{8} = 0.95339 \dots

u_{9} = 0.30154 \dots : Δ u_{9} = 0.61768 \dots

f (u_{8}) = (- 0.31613 \dots)^{2} - 0.83928 \dots (- 0.31613 \dots) + 0.54368 \dots = 0.90896 \dots

f^{^{'}} (u_{8}) = 2 (- 0.31613 \dots) - 0.83928 \dots = - 1.47156 \dots

u_{9} = 0.30154 \dots

Δ u_{9} = ∣ 0.30154 \dots - (- 0.31613 \dots) ∣ = 0.61768 \dots

Δ u_{9} = 0.61768 \dots

u_{10} = 1.91692 \dots : Δ u_{10} = 1.61537 \dots

f (u_{9}) = 0.30154 \dots^{2} - 0.83928 \dots \cdot 0.30154 \dots + 0.54368 \dots = 0.38153 \dots

f^{^{'}} (u_{9}) = 2 \cdot 0.30154 \dots - 0.83928 \dots = - 0.23619 \dots

u_{10} = 1.91692 \dots

Δ u_{10} = ∣ 1.91692 \dots - 0.30154 \dots ∣ = 1.61537 \dots

Δ u_{10} = 1.61537 \dots

u_{11} = 1.04552 \dots : Δ u_{11} = 0.87139 \dots

f (u_{10}) = 1.91692 \dots^{2} - 0.83928 \dots \cdot 1.91692 \dots + 0.54368 \dots = 2.60942 \dots

f^{^{'}} (u_{10}) = 2 \cdot 1.91692 \dots - 0.83928 \dots = 2.99455 \dots

u_{11} = 1.04552 \dots

Δ u_{11} = ∣ 1.04552 \dots - 1.91692 \dots ∣ = 0.87139 \dots

Δ u_{11} = 0.87139 \dots

u_{12} = 0.43893 \dots : Δ u_{12} = 0.60659 \dots

f (u_{11}) = 1.04552 \dots^{2} - 0.83928 \dots \cdot 1.04552 \dots + 0.54368 \dots = 0.75932 \dots

f^{^{'}} (u_{11}) = 2 \cdot 1.04552 \dots - 0.83928 \dots = 1.25177 \dots

u_{12} = 0.43893 \dots

Δ u_{12} = ∣ 0.43893 \dots - 1.04552 \dots ∣ = 0.60659 \dots

Δ u_{12} = 0.60659 \dots

u_{13} = - 9.09911 \dots : Δ u_{13} = 9.53804 \dots

f (u_{12}) = 0.43893 \dots^{2} - 0.83928 \dots \cdot 0.43893 \dots + 0.54368 \dots = 0.36796 \dots

f^{^{'}} (u_{12}) = 2 \cdot 0.43893 \dots - 0.83928 \dots = 0.03857 \dots

u_{13} = - 9.09911 \dots

Δ u_{13} = ∣ - 9.09911 \dots - 0.43893 \dots ∣ = 9.53804 \dots

Δ u_{13} = 9.53804 \dots

لا يمكن إيجاد حلّ

الحل للمعادلة هو

u \approx - 1.83928 \dots

u \approx - 1.83928 \dots

افحص الإجبات

جد نقاط غير معرّفة:

u = 1, u = - 1

وقم بمساواتها لصفر

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

خذ المقامات في

1 - u^{2} = 0

حلّ:

u = 1, u = - 1

1 - u^{2} = 0

انقل

1

إلى الجانب الأيمن

1 - u^{2} = 0

من الطرفين

1

اطرح

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

بسّط

- u^{2} = - 1

- u^{2} = - 1

- 1

اقسم الطرفين على

- u^{2} = - 1

- 1

اقسم الطرفين على

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

بسّط

u^{2} = 1

u^{2} = 1

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

الحلول هي

x^{2} = f (a)

لـ

u = 1, u = - 1

1 = 1

1

1 = 1

:فعْل قانون الجذور

= 1

- 1 = - 1

- 1

1 = 1

:فعْل قانون الجذور

1 = 1

= - 1

u = 1, u = - 1

النقاط التالية غير معرّفة

u = 1, u = - 1

ضمّ النقاط غير المعرّفة مع الحلول

u \approx - 1.83928 \dots

u = cos (x)

استبدل مجددًا

cos (x) \approx - 1.83928 \dots

cos (x) \approx - 1.83928 \dots

cos (x) = - 1.83928 \dots :

لا يوجد حلّ

cos (x) = - 1.83928 \dots

- 1 \leq cos (x) \leq 1

لا يوجد حلّ

وحّد الحلول

x \in R لا يوجد حلّ لـ

رسم

أمثلة شائعة

6arccos(5x)=5pi

6 arccos (5 x) = 5 π

cos(2θ)=-1/2 ,0<= θ<= 2pi

cos (2 θ) = - \frac{1}{2}, 0 \leq θ \leq 2 π

sin(2x)= 2/5

sin (2 x) = \frac{2}{5}

sin(2x)+cos(x)=0,0<= x<= 2pi

sin (2 x) + cos (x) = 0, 0 \leq x \leq 2 π

cos(θ)=-0.5

cos (θ) = - 0.5