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csc^3(x)-4csc(x)=3cot^2(x)

الحلّ

csc^{3} (x) - 4 csc (x) = 3 cot^{2} (x)

الحلّ

x = - 0.79731 \dots + 2 πn, x = π + 0.79731 \dots + 2 πn, x = 0.26356 \dots + 2 πn, x = π - 0.26356 \dots + 2 πn

درجات

x = - 45.68261 \dots^{\circ} + 36 0^{\circ} n, x = 225.68261 \dots^{\circ} + 36 0^{\circ} n, x = 15.10095 \dots^{\circ} + 36 0^{\circ} n, x = 164.89904 \dots^{\circ} + 36 0^{\circ} n

خطوات الحلّ

csc^{3} (x) - 4 csc (x) = 3 cot^{2} (x)

من الطرفين

3 cot^{2} (x)

اطرح

csc^{3} (x) - 4 csc (x) - 3 cot^{2} (x) = 0

Rewrite using trig identities

csc^{3} (x) - 3 cot^{2} (x) - 4 csc (x)

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

:فعّل نطريّة فيتاغوروس

cot^{2} (x) = csc^{2} (x) - 1

= csc^{3} (x) - 3 (csc^{2} (x) - 1) - 4 csc (x)

csc^{3} (x) - (- 1 + csc^{2} (x)) \cdot 3 - 4 csc (x) = 0

بالاستعانة بطريقة التعويض

csc^{3} (x) - (- 1 + csc^{2} (x)) \cdot 3 - 4 csc (x) = 0

csc (x) = u :

على افتراض أنّ

u^{3} - (- 1 + u^{2}) \cdot 3 - 4 u = 0

u^{3} - (- 1 + u^{2}) \cdot 3 - 4 u = 0 : u \approx 0.55919 \dots, u \approx - 1.39766 \dots, u \approx 3.83846 \dots

u^{3} - (- 1 + u^{2}) \cdot 3 - 4 u = 0

u^{3} - (- 1 + u^{2}) \cdot 3 - 4 u

وسّع:

u^{3} + 3 - 3 u^{2} - 4 u

u^{3} - (- 1 + u^{2}) \cdot 3 - 4 u

= u^{3} - 3 (- 1 + u^{2}) - 4 u

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

وسٌع:

3 - 3 u^{2}

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

a (b + c) = ab + a c

: افتح أقواس بالاستعانة بـ

a = - 3, b = - 1, c = u^{2}

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

فعّل قوانين سالب-موجب

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

= 3 \cdot 1 - 3 u^{2}

3 \cdot 1 = 3 :

اضرب الأعداد

= 3 - 3 u^{2}

= u^{3} + 3 - 3 u^{2} - 4 u

u^{3} + 3 - 3 u^{2} - 4 u = 0

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

اكتب بالصورة الاعتياديّة

u^{3} - 3 u^{2} - 4 u + 3 = 0

بطريقة نيوتون ريبسون

u^{3} - 3 u^{2} - 4 u + 3 = 0

جدّ حلًا لـ:

u \approx 0.55919 \dots

u^{3} - 3 u^{2} - 4 u + 3 = 0

تعريف تقريب نيوتن-ريبسون

f (u) = u^{3} - 3 u^{2} - 4 u + 3

f^{^{'}} (u)

جد:

3 u^{2} - 6 u - 4

\frac{d}{d u} (u^{3} - 3 u^{2} - 4 u + 3)

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

:استعمل قانون الجمع

= \frac{d}{d u} (u^{3}) - \frac{d}{d u} (3 u^{2}) - \frac{d}{d u} (4 u) + \frac{d}{d u} (3)

\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} (3 u^{2}) = 6 u

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

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

:استخرج الثابت

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

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

:استعمل قانون الأسس

= 3 \cdot 2 u^{2 - 1}

بسّط

= 6 u

\frac{d}{d u} (4 u) = 4

\frac{d}{d u} (4 u)

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

:استخرج الثابت

= 4 \frac{d u}{d u}

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

:استعمل المشتقة الأساسية

= 4 \cdot 1

بسّط

= 4

\frac{d}{d u} (3) = 0

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

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

:مشتقة الثابت

= 0

= 3 u^{2} - 6 u - 4 + 0

بسّط

= 3 u^{2} - 6 u - 4

u_{0} = 1

استبدل

Δ u_{n + 1} < 0.000001

حتّى

u_{n + 1}

احسب

u_{1} = 0.57142 \dots : Δ u_{1} = 0.42857 \dots

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

f^{^{'}} (u_{0}) = 3 \cdot 1^{2} - 6 \cdot 1 - 4 = - 7

u_{1} = 0.57142 \dots

Δ u_{1} = ∣ 0.57142 \dots - 1 ∣ = 0.42857 \dots

Δ u_{1} = 0.42857 \dots

u_{2} = 0.55922 \dots : Δ u_{2} = 0.01220 \dots

f (u_{1}) = 0.57142 \dots^{3} - 3 \cdot 0.57142 \dots^{2} - 4 \cdot 0.57142 \dots + 3 = - 0.07871 \dots

f^{^{'}} (u_{1}) = 3 \cdot 0.57142 \dots^{2} - 6 \cdot 0.57142 \dots - 4 = - 6.44897 \dots

u_{2} = 0.55922 \dots

Δ u_{2} = ∣ 0.55922 \dots - 0.57142 \dots ∣ = 0.01220 \dots

Δ u_{2} = 0.01220 \dots

u_{3} = 0.55919 \dots : Δ u_{3} = 0.00003 \dots

f (u_{2}) = 0.55922 \dots^{3} - 3 \cdot 0.55922 \dots^{2} - 4 \cdot 0.55922 \dots + 3 = - 0.00019 \dots

f^{^{'}} (u_{2}) = 3 \cdot 0.55922 \dots^{2} - 6 \cdot 0.55922 \dots - 4 = - 6.41714 \dots

u_{3} = 0.55919 \dots

Δ u_{3} = ∣ 0.55919 \dots - 0.55922 \dots ∣ = 0.00003 \dots

Δ u_{3} = 0.00003 \dots

u_{4} = 0.55919 \dots : Δ u_{4} = 1.87129 E - 10

f (u_{3}) = 0.55919 \dots^{3} - 3 \cdot 0.55919 \dots^{2} - 4 \cdot 0.55919 \dots + 3 = - 1.20082 E - 9

f^{^{'}} (u_{3}) = 3 \cdot 0.55919 \dots^{2} - 6 \cdot 0.55919 \dots - 4 = - 6.41706 \dots

u_{4} = 0.55919 \dots

Δ u_{4} = ∣ 0.55919 \dots - 0.55919 \dots ∣ = 1.87129 E - 10

Δ u_{4} = 1.87129 E - 10

u \approx 0.55919 \dots

فعّل القسمة الطويلة:

\frac{u ^{3} - 3 u ^{2} - 4 u + 3}{u - 0.55919 \dots} = u^{2} - 2.44080 \dots u - 5.36488 \dots

u^{2} - 2.44080 \dots u - 5.36488 \dots \approx 0

بطريقة نيوتون ريبسون

u^{2} - 2.44080 \dots u - 5.36488 \dots = 0

جدّ حلًا لـ:

u \approx - 1.39766 \dots

u^{2} - 2.44080 \dots u - 5.36488 \dots = 0

تعريف تقريب نيوتن-ريبسون

f (u) = u^{2} - 2.44080 \dots u - 5.36488 \dots

f^{^{'}} (u)

جد:

2 u - 2.44080 \dots

\frac{d}{d u} (u^{2} - 2.44080 \dots u - 5.36488 \dots)

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

:استعمل قانون الجمع

= \frac{d}{d u} (u^{2}) - \frac{d}{d u} (2.44080 \dots u) - \frac{d}{d u} (5.36488 \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} (2.44080 \dots u) = 2.44080 \dots

\frac{d}{d u} (2.44080 \dots u)

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

:استخرج الثابت

= 2.44080 \dots \frac{d u}{d u}

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

:استعمل المشتقة الأساسية

= 2.44080 \dots \cdot 1

بسّط

= 2.44080 \dots

\frac{d}{d u} (5.36488 \dots) = 0

\frac{d}{d u} (5.36488 \dots)

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

:مشتقة الثابت

= 0

= 2 u - 2.44080 \dots - 0

بسّط

= 2 u - 2.44080 \dots

u_{0} = - 2

استبدل

Δ u_{n + 1} < 0.000001

حتّى

u_{n + 1}

احسب

u_{1} = - 1.45399 \dots : Δ u_{1} = 0.54600 \dots

f (u_{0}) = (- 2)^{2} - 2.44080 \dots (- 2) - 5.36488 \dots = 3.51673 \dots

f^{^{'}} (u_{0}) = 2 (- 2) - 2.44080 \dots = - 6.44080 \dots

u_{1} = - 1.45399 \dots

Δ u_{1} = ∣ - 1.45399 \dots - (- 2) ∣ = 0.54600 \dots

Δ u_{1} = 0.54600 \dots

u_{2} = - 1.39825 \dots : Δ u_{2} = 0.05573 \dots

f (u_{1}) = (- 1.45399 \dots)^{2} - 2.44080 \dots (- 1.45399 \dots) - 5.36488 \dots = 0.29812 \dots

f^{^{'}} (u_{1}) = 2 (- 1.45399 \dots) - 2.44080 \dots = - 5.34879 \dots

u_{2} = - 1.39825 \dots

Δ u_{2} = ∣ - 1.39825 \dots - (- 1.45399 \dots) ∣ = 0.05573 \dots

Δ u_{2} = 0.05573 \dots

u_{3} = - 1.39766 \dots : Δ u_{3} = 0.00059 \dots

f (u_{2}) = (- 1.39825 \dots)^{2} - 2.44080 \dots (- 1.39825 \dots) - 5.36488 \dots = 0.00310 \dots

f^{^{'}} (u_{2}) = 2 (- 1.39825 \dots) - 2.44080 \dots = - 5.23731 \dots

u_{3} = - 1.39766 \dots

Δ u_{3} = ∣ - 1.39766 \dots - (- 1.39825 \dots) ∣ = 0.00059 \dots

Δ u_{3} = 0.00059 \dots

u_{4} = - 1.39766 \dots : Δ u_{4} = 6.7196 E - 8

f (u_{3}) = (- 1.39766 \dots)^{2} - 2.44080 \dots (- 1.39766 \dots) - 5.36488 \dots = 3.51847 E - 7

f^{^{'}} (u_{3}) = 2 (- 1.39766 \dots) - 2.44080 \dots = - 5.23613 \dots

u_{4} = - 1.39766 \dots

Δ u_{4} = ∣ - 1.39766 \dots - (- 1.39766 \dots) ∣ = 6.7196 E - 8

Δ u_{4} = 6.7196 E - 8

u \approx - 1.39766 \dots

فعّل القسمة الطويلة:

\frac{u ^{2} - 2.44080 \dots u - 5.36488 \dots}{u + 1.39766 \dots} = u - 3.83846 \dots

u - 3.83846 \dots \approx 0

u \approx 3.83846 \dots

The solutions are

u \approx 0.55919 \dots, u \approx - 1.39766 \dots, u \approx 3.83846 \dots

u = csc (x)

استبدل مجددًا

csc (x) \approx 0.55919 \dots, csc (x) \approx - 1.39766 \dots, csc (x) \approx 3.83846 \dots

csc (x) \approx 0.55919 \dots, csc (x) \approx - 1.39766 \dots, csc (x) \approx 3.83846 \dots

csc (x) = 0.55919 \dots :

لا يوجد حلّ

csc (x) = 0.55919 \dots

csc (x) \leq - 1 or csc (x) \geq 1

لا يوجد حلّ

csc (x) = - 1.39766 \dots : x = arccsc (- 1.39766 \dots) + 2 πn, x = π + arccsc (1.39766 \dots) + 2 πn

csc (x) = - 1.39766 \dots

Apply trig inverse properties

csc (x) = - 1.39766 \dots

csc (x) = - 1.39766 \dots :

حلول عامّة لـ

csc (x) = - a \Rightarrow x = arccsc (- a) + 2 πn, x = π + arccsc (a) + 2 πn

x = arccsc (- 1.39766 \dots) + 2 πn, x = π + arccsc (1.39766 \dots) + 2 πn

x = arccsc (- 1.39766 \dots) + 2 πn, x = π + arccsc (1.39766 \dots) + 2 πn

csc (x) = 3.83846 \dots : x = arccsc (3.83846 \dots) + 2 πn, x = π - arccsc (3.83846 \dots) + 2 πn

csc (x) = 3.83846 \dots

Apply trig inverse properties

csc (x) = 3.83846 \dots

csc (x) = 3.83846 \dots :

حلول عامّة لـ

csc (x) = a \Rightarrow x = arccsc (a) + 2 πn, x = π - arccsc (a) + 2 πn

x = arccsc (3.83846 \dots) + 2 πn, x = π - arccsc (3.83846 \dots) + 2 πn

x = arccsc (3.83846 \dots) + 2 πn, x = π - arccsc (3.83846 \dots) + 2 πn

وحّد الحلول

x = arccsc (- 1.39766 \dots) + 2 πn, x = π + arccsc (1.39766 \dots) + 2 πn, x = arccsc (3.83846 \dots) + 2 πn, x = π - arccsc (3.83846 \dots) + 2 πn

أظهر الحلّ بالتمثيل العشريّ

x = - 0.79731 \dots + 2 πn, x = π + 0.79731 \dots + 2 πn, x = 0.26356 \dots + 2 πn, x = π - 0.26356 \dots + 2 πn

أمثلة شائعة

1+tan^2(θ)=4tan^2(θ)

1 + tan^{2} (θ) = 4 tan^{2} (θ)

(tan(θ)+1)(cos(θ)+1)(sin(θ))=0

(tan (θ) + 1) (cos (θ) + 1) (sin (θ)) = 0

6sin(2x)-3=0

6 sin (2 x) - 3 = 0

4cos^2(x)-sin(x)=1

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

0.6=cos(4x)

0.6 = cos (4 x)