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شائع علم المثلثات >

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

الحلّ

cos^{2} (x) + cos^{2} (3 x) = 1

الحلّ

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn, x = 0.39269 \dots + 2 πn, x = 2 π - 0.39269 \dots + 2 πn, x = 2.74889 \dots + 2 πn, x = - 2.74889 \dots + 2 πn, x = 1.17809 \dots + 2 πn, x = 2 π - 1.17809 \dots + 2 πn, x = 1.96349 \dots + 2 πn, x = - 1.96349 \dots + 2 πn

درجات

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = - 13 5^{\circ} + 36 0^{\circ} n, x = 22. 5^{\circ} + 36 0^{\circ} n, x = 337. 5^{\circ} + 36 0^{\circ} n, x = 157. 5^{\circ} + 36 0^{\circ} n, x = - 157. 5^{\circ} + 36 0^{\circ} n, x = 67. 5^{\circ} + 36 0^{\circ} n, x = 292. 5^{\circ} + 36 0^{\circ} n, x = 112. 5^{\circ} + 36 0^{\circ} n, x = - 112. 5^{\circ} + 36 0^{\circ} n

خطوات الحلّ

cos^{2} (x) + cos^{2} (3 x) = 1

من الطرفين

1

اطرح

cos^{2} (x) + cos^{2} (3 x) - 1 = 0

Rewrite using trig identities

- 1 + cos^{2} (3 x) + cos^{2} (x)

cos (3 x) = 4 cos^{3} (x) - 3 cos (x)

cos (3 x)

Rewrite using trig identities

cos (3 x)

أعد الكتابة كـ

= cos (2 x + x)

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

:فعّل متطابقة الجمع لزوايا

= cos (2 x) cos (x) - sin (2 x) sin (x)

sin (2 x) = 2 sin (x) cos (x)

:فعّل متطابقة الزاوية المضاعفة

= cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

بسّط:

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

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

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

2 sin (x) cos (x) sin (x)

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

:فعّل قانون القوى

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

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

1 + 1 = 2 :

اجمع الأعداد

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

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

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

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

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

:فعّل متطابقة الزاوية المضاعفة

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

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

:فعّل نطريّة فيتاغوروس

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

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

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

وسٌع:

4 cos^{3} (x) - 3 cos (x)

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

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

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

وسٌع:

2 cos^{3} (x) - cos (x)

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

a (b - c) = ab - a c

: افتح أقواس بالاستعانة بـ

a = cos (x), b = 2 cos^{2} (x), c = 1

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

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

2 cos^{2} (x) cos (x) - 1 \cdot cos (x)

بسّط:

2 cos^{3} (x) - cos (x)

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

2 cos^{2} (x) cos (x) = 2 cos^{3} (x)

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

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

:فعّل قانون القوى

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

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

2 + 1 = 3 :

اجمع الأعداد

= 2 cos^{3} (x)

1 \cdot cos (x) = cos (x)

1 cos (x)

1 \cdot cos (x) = cos (x) :

اضرب

= cos (x)

= 2 cos^{3} (x) - cos (x)

= 2 cos^{3} (x) - cos (x)

= 2 cos^{3} (x) - cos (x) - 2 (1 - cos^{2} (x)) cos (x)

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

وسٌع:

- 2 cos (x) + 2 cos^{3} (x)

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

a (b - c) = ab - a c

: افتح أقواس بالاستعانة بـ

a = - 2 cos (x), b = 1, c = cos^{2} (x)

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

فعّل قوانين سالب-موجب

- (- a) = a

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

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

بسّط:

- 2 cos (x) + 2 cos^{3} (x)

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

2 \cdot 1 \cdot cos (x) = 2 cos (x)

2 \cdot 1 cos (x)

2 \cdot 1 = 2 :

اضرب الأعداد

= 2 cos (x)

2 cos^{2} (x) cos (x) = 2 cos^{3} (x)

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

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

:فعّل قانون القوى

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

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

2 + 1 = 3 :

اجمع الأعداد

= 2 cos^{3} (x)

= - 2 cos (x) + 2 cos^{3} (x)

= - 2 cos (x) + 2 cos^{3} (x)

= 2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x)

2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x)

بسّط:

4 cos^{3} (x) - 3 cos (x)

2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x)

جمّع التعابير المتشابهة

= 2 cos^{3} (x) + 2 cos^{3} (x) - cos (x) - 2 cos (x)

2 cos^{3} (x) + 2 cos^{3} (x) = 4 cos^{3} (x) :

اجمع العناصر المتشابهة

= 4 cos^{3} (x) - cos (x) - 2 cos (x)

- cos (x) - 2 cos (x) = - 3 cos (x) :

اجمع العناصر المتشابهة

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

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

- 1 + (4 cos^{3} (x) - 3 cos (x))^{2} + cos^{2} (x)

بسّط:

- 1 + 16 cos^{6} (x) - 24 cos^{4} (x) + 10 cos^{2} (x)

- 1 + (4 cos^{3} (x) - 3 cos (x))^{2} + cos^{2} (x)

(4 cos^{3} (x) - 3 cos (x))^{2} : 16 cos^{6} (x) - 24 cos^{4} (x) + 9 cos^{2} (x)

(a - b)^{2} = a^{2} - 2 ab + b^{2}

:فعّل صيغة الضرب المختصر

a = 4 cos^{3} (x), b = 3 cos (x)

= (4 cos^{3} (x))^{2} - 2 \cdot 4 cos^{3} (x) \cdot 3 cos (x) + (3 cos (x))^{2}

(4 cos^{3} (x))^{2} - 2 \cdot 4 cos^{3} (x) \cdot 3 cos (x) + (3 cos (x))^{2}

بسّط:

16 cos^{6} (x) - 24 cos^{4} (x) + 9 cos^{2} (x)

(4 cos^{3} (x))^{2} - 2 \cdot 4 cos^{3} (x) \cdot 3 cos (x) + (3 cos (x))^{2}

(4 cos^{3} (x))^{2} = 16 cos^{6} (x)

(4 cos^{3} (x))^{2}

(a \cdot b)^{n} = a^{n} b^{n}

:فعّل قانون القوى

= 4^{2} (cos^{3} (x))^{2}

(cos^{3} (x))^{2} : cos^{6} (x)

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

:فعّل قانون القوى

= cos^{3 \cdot 2} (x)

3 \cdot 2 = 6 :

اضرب الأعداد

= cos^{6} (x)

= 4^{2} cos^{6} (x)

4^{2} = 16

= 16 cos^{6} (x)

2 \cdot 4 cos^{3} (x) \cdot 3 cos (x) = 24 cos^{4} (x)

2 \cdot 4 cos^{3} (x) \cdot 3 cos (x)

2 \cdot 4 \cdot 3 = 24 :

اضرب الأعداد

= 24 cos^{3} (x) cos (x)

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

:فعّل قانون القوى

cos^{3} (x) cos (x) = cos^{3 + 1} (x)

= 24 cos^{3 + 1} (x)

3 + 1 = 4 :

اجمع الأعداد

= 24 cos^{4} (x)

(3 cos (x))^{2} = 9 cos^{2} (x)

(3 cos (x))^{2}

(a \cdot b)^{n} = a^{n} b^{n}

:فعّل قانون القوى

= 3^{2} cos^{2} (x)

3^{2} = 9

= 9 cos^{2} (x)

= 16 cos^{6} (x) - 24 cos^{4} (x) + 9 cos^{2} (x)

= 16 cos^{6} (x) - 24 cos^{4} (x) + 9 cos^{2} (x)

= - 1 + 16 cos^{6} (x) - 24 cos^{4} (x) + 9 cos^{2} (x) + cos^{2} (x)

9 cos^{2} (x) + cos^{2} (x) = 10 cos^{2} (x) :

اجمع العناصر المتشابهة

= - 1 + 16 cos^{6} (x) - 24 cos^{4} (x) + 10 cos^{2} (x)

= - 1 + 16 cos^{6} (x) - 24 cos^{4} (x) + 10 cos^{2} (x)

- 1 + 10 cos^{2} (x) + 16 cos^{6} (x) - 24 cos^{4} (x) = 0

بالاستعانة بطريقة التعويض

- 1 + 10 cos^{2} (x) + 16 cos^{6} (x) - 24 cos^{4} (x) = 0

cos (x) = u :

على افتراض أنّ

- 1 + 10 u^{2} + 16 u^{6} - 24 u^{4} = 0

- 1 + 10 u^{2} + 16 u^{6} - 24 u^{4} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}, u = \frac{2 + 2}{2}, u = - \frac{2 + 2}{2}, u = \frac{2 - 2}{2}, u = - \frac{2 - 2}{2}

- 1 + 10 u^{2} + 16 u^{6} - 24 u^{4} = 0

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

اكتب بالصورة الاعتياديّة

16 u^{6} - 24 u^{4} + 10 u^{2} - 1 = 0

v^{3} = u^{6}

وكذلك

v = u^{2}, v^{2} = u^{4}

اكتب المعادلة مجددًا، بحيث أنّ

16 v^{3} - 24 v^{2} + 10 v - 1 = 0

16 v^{3} - 24 v^{2} + 10 v - 1 = 0

حلّ:

v = \frac{1}{2}, v = \frac{2 + 2}{4}, v = \frac{2 - 2}{4}

16 v^{3} - 24 v^{2} + 10 v - 1 = 0

16 v^{3} - 24 v^{2} + 10 v - 1

حلّل إلى عوامل:

(2 v - 1) (8 v^{2} - 8 v + 1)

16 v^{3} - 24 v^{2} + 10 v - 1

استعمل نظريّة الجذر الكسريّ

2 v - 1

هو جذر للتعبير، إذًا فلتخرج

\pm \frac{1}{1 , 2 , 4 , 8 , 16}

\frac{1}{2}

لذلك، افحص الأعداد الكسريّة التالية

a_{n} : 1, 2, 4, 8, 16

القواسم لـ

a_{0} : 1,

القواسم لـ

a_{0} = 1, a_{n} = 16

= (2 v - 1) \frac{16 v ^{3} - 24 v ^{2} + 10 v - 1}{2 v - 1}

\frac{16 v ^{3} - 24 v ^{2} + 10 v - 1}{2 v - 1} = 8 v^{2} - 8 v + 1

\frac{16 v ^{3} - 24 v ^{2} + 10 v - 1}{2 v - 1}

\frac{16 v ^{3} - 24 v ^{2} + 10 v - 1}{2 v - 1}

اقسم:

\frac{16 v ^{3} - 24 v ^{2} + 10 v - 1}{2 v - 1} = 8 v^{2} + \frac{- 16 v ^{2} + 10 v - 1}{2 v - 1}

16 v^{3} - 24 v^{2} + 10 v - 1

اقسم المعامل الرئيس للبسط

\frac{16 v ^{3}}{2 v} = 8 v^{2} : 2 v - 1

والمقام

Quotient = 8 v^{2}

16 v^{3} - 8 v^{2} : 8 v^{2}

بـ

2 v - 1

اضرب للحصول على باقٍ جديد

16 v^{3} - 24 v^{2} + 10 v - 1

من

16 v^{3} - 8 v^{2}

اطرح

باقي = - 16 v^{2} + 10 v - 1

لذلك

\frac{16 v ^{3} - 24 v ^{2} + 10 v - 1}{2 v - 1} = 8 v^{2} + \frac{- 16 v ^{2} + 10 v - 1}{2 v - 1}

= 8 v^{2} + \frac{- 16 v ^{2} + 10 v - 1}{2 v - 1}

\frac{- 16 v ^{2} + 10 v - 1}{2 v - 1}

اقسم:

\frac{- 16 v ^{2} + 10 v - 1}{2 v - 1} = - 8 v + \frac{2 v - 1}{2 v - 1}

- 16 v^{2} + 10 v - 1

اقسم المعامل الرئيس للبسط

\frac{- 16 v ^{2}}{2 v} = - 8 v : 2 v - 1

والمقام

Quotient = - 8 v

- 16 v^{2} + 8 v : - 8 v

بـ

2 v - 1

اضرب للحصول على باقٍ جديد

- 16 v^{2} + 10 v - 1

من

- 16 v^{2} + 8 v

اطرح

باقي = 2 v - 1

لذلك

\frac{- 16 v ^{2} + 10 v - 1}{2 v - 1} = - 8 v + \frac{2 v - 1}{2 v - 1}

= 8 v^{2} - 8 v + \frac{2 v - 1}{2 v - 1}

\frac{2 v - 1}{2 v - 1}

اقسم:

\frac{2 v - 1}{2 v - 1} = 1

2 v - 1

اقسم المعامل الرئيس للبسط

\frac{2 v}{2 v} = 1 : 2 v - 1

والمقام

Quotient = 1

2 v - 1 : 1

بـ

2 v - 1

اضرب للحصول على باقٍ جديد

2 v - 1

من

2 v - 1

اطرح

باقي = 0

لذلك

\frac{2 v - 1}{2 v - 1} = 1

= 8 v^{2} - 8 v + 1

= (2 v - 1) (8 v^{2} - 8 v + 1)

(2 v - 1) (8 v^{2} - 8 v + 1) = 0

حلّ عن طريق مساواة العوامل لصفر

2 v - 1 = 0 or 8 v^{2} - 8 v + 1 = 0

2 v - 1 = 0

حلّ:

v = \frac{1}{2}

2 v - 1 = 0

انقل

1

إلى الجانب الأيمن

2 v - 1 = 0

للطرفين

1

أضف

2 v - 1 + 1 = 0 + 1

بسّط

2 v = 1

2 v = 1

2

اقسم الطرفين على

2 v = 1

2

اقسم الطرفين على

\frac{2 v}{2} = \frac{1}{2}

بسّط

v = \frac{1}{2}

v = \frac{1}{2}

8 v^{2} - 8 v + 1 = 0

حلّ:

v = \frac{2 + 2}{4}, v = \frac{2 - 2}{4}

8 v^{2} - 8 v + 1 = 0

حلّ بالاستعانة بالصيغة التربيعيّة

8 v^{2} - 8 v + 1 = 0

الصيغة لحلّ المعادلة التربيعيّة

a = 8, b = - 8, c = 1

لـ

v_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

v_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

(- 8)^{2} - 4 \cdot 8 \cdot 1 = 42

(- 8)^{2} - 4 \cdot 8 \cdot 1

زوجيّ n

إذا تحقّق أنّ

, (- a)^{n} = a^{n}

:فعّل قانون القوى

(- 8)^{2} = 8^{2}

= 8^{2} - 4 \cdot 8 \cdot 1

4 \cdot 8 \cdot 1 = 32 :

اضرب الأعداد

= 8^{2} - 32

8^{2} = 64

= 64 - 32

64 - 32 = 32 :

اطرح الأعداد

= 32

32

تحليل لعوامل أوّليّة لـ:

2^{5}

32

32 = 16 \cdot 2, 2

ينقسم على

32

= 2 \cdot 16

16 = 8 \cdot 2, 2

ينقسم على

16

= 2 \cdot 2 \cdot 8

8 = 4 \cdot 2, 2

ينقسم على

8

= 2 \cdot 2 \cdot 2 \cdot 4

4 = 2 \cdot 2, 2

ينقسم على

4

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

هو عدد أوّليّ لذلك تحليل آخر لعوامل غير ممكن

2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 2^{5}

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

:فعّل قانون القوى

= 2^{4} \cdot 2

n ab = n a n b

:فعْل قانون الجذور

= 2 2^{4}

n a^{m} = a^{\frac{m}{n}}

:فعْل قانون الجذور

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 2

بسّط

= 42

v_{1, 2} = \frac{- ( - 8 ) \pm 4 2}{2 \cdot 8}

Separate the solutions

v_{1} = \frac{- ( - 8 ) + 4 2}{2 \cdot 8}, v_{2} = \frac{- ( - 8 ) - 4 2}{2 \cdot 8}

v = \frac{- ( - 8 ) + 4 2}{2 \cdot 8} : \frac{2 + 2}{4}

\frac{- ( - 8 ) + 4 2}{2 \cdot 8}

- (- a) = a

فعّل القانون

= \frac{8 + 4 2}{2 \cdot 8}

2 \cdot 8 = 16 :

اضرب الأعداد

= \frac{8 + 4 2}{16}

8 + 42

حلل إلى عوامل:

4 (2 + 2)

8 + 42

أعد الكتابة كـ

= 4 \cdot 2 + 42

4

قم باخراج العامل المشترك

= 4 (2 + 2)

= \frac{4 ( 2 + 2 )}{16}

4 :

إلغ العوامل المشتركة

= \frac{2 + 2}{4}

v = \frac{- ( - 8 ) - 4 2}{2 \cdot 8} : \frac{2 - 2}{4}

\frac{- ( - 8 ) - 4 2}{2 \cdot 8}

- (- a) = a

فعّل القانون

= \frac{8 - 4 2}{2 \cdot 8}

2 \cdot 8 = 16 :

اضرب الأعداد

= \frac{8 - 4 2}{16}

8 - 42

حلل إلى عوامل:

4 (2 - 2)

8 - 42

أعد الكتابة كـ

= 4 \cdot 2 - 42

4

قم باخراج العامل المشترك

= 4 (2 - 2)

= \frac{4 ( 2 - 2 )}{16}

4 :

إلغ العوامل المشتركة

= \frac{2 - 2}{4}

حلول المعادلة التربيعيّة هي

v = \frac{2 + 2}{4}, v = \frac{2 - 2}{4}

The solutions are

v = \frac{1}{2}, v = \frac{2 + 2}{4}, v = \frac{2 - 2}{4}

v = \frac{1}{2}, v = \frac{2 + 2}{4}, v = \frac{2 - 2}{4}

Substitute back

v = u^{2},

solve for

u

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

حلّ:

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

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

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

الحلول هي

x^{2} = f (a)

لـ

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

u^{2} = \frac{2 + 2}{4}

حلّ:

u = \frac{2 + 2}{2}, u = - \frac{2 + 2}{2}

u^{2} = \frac{2 + 2}{4}

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

الحلول هي

x^{2} = f (a)

لـ

u = \frac{2 + 2}{4}, u = - \frac{2 + 2}{4}

\frac{2 + 2}{4} = \frac{2 + 2}{2}

\frac{2 + 2}{4}

a \geq 0, b \geq 0

بافتراض أنّ

n \frac{a}{b} = \frac{n a}{n b}

:فعّل قانون الجذور

= \frac{2 + 2}{4}

4 = 2

4

4 = 2^{2} :

حلّل العدد لعوامله أوّليّة

= 2^{2}

n a^{n} = a

:فعْل قانون الجذور

2^{2} = 2

= 2

= \frac{2 + 2}{2}

- \frac{2 + 2}{4} = - \frac{2 + 2}{2}

- \frac{2 + 2}{4}

\frac{2 + 2}{4}

بسّط:

\frac{2 + 2}{2}

\frac{2 + 2}{4}

a \geq 0, b \geq 0

بافتراض أنّ

n \frac{a}{b} = \frac{n a}{n b}

:فعّل قانون الجذور

= \frac{2 + 2}{4}

4 = 2

4

4 = 2^{2} :

حلّل العدد لعوامله أوّليّة

= 2^{2}

n a^{n} = a

:فعْل قانون الجذور

2^{2} = 2

= 2

= \frac{2 + 2}{2}

= - \frac{2 + 2}{2}

u = \frac{2 + 2}{2}, u = - \frac{2 + 2}{2}

u^{2} = \frac{2 - 2}{4}

حلّ:

u = \frac{2 - 2}{2}, u = - \frac{2 - 2}{2}

u^{2} = \frac{2 - 2}{4}

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

الحلول هي

x^{2} = f (a)

لـ

u = \frac{2 - 2}{4}, u = - \frac{2 - 2}{4}

\frac{2 - 2}{4} = \frac{2 - 2}{2}

\frac{2 - 2}{4}

a \geq 0, b \geq 0

بافتراض أنّ

n \frac{a}{b} = \frac{n a}{n b}

:فعّل قانون الجذور

= \frac{2 - 2}{4}

4 = 2

4

4 = 2^{2} :

حلّل العدد لعوامله أوّليّة

= 2^{2}

n a^{n} = a

:فعْل قانون الجذور

2^{2} = 2

= 2

= \frac{2 - 2}{2}

- \frac{2 - 2}{4} = - \frac{2 - 2}{2}

- \frac{2 - 2}{4}

\frac{2 - 2}{4}

بسّط:

\frac{2 - 2}{2}

\frac{2 - 2}{4}

a \geq 0, b \geq 0

بافتراض أنّ

n \frac{a}{b} = \frac{n a}{n b}

:فعّل قانون الجذور

= \frac{2 - 2}{4}

4 = 2

4

4 = 2^{2} :

حلّل العدد لعوامله أوّليّة

= 2^{2}

n a^{n} = a

:فعْل قانون الجذور

2^{2} = 2

= 2

= \frac{2 - 2}{2}

= - \frac{2 - 2}{2}

u = \frac{2 - 2}{2}, u = - \frac{2 - 2}{2}

The solutions are

u = \frac{1}{2}, u = - \frac{1}{2}, u = \frac{2 + 2}{2}, u = - \frac{2 + 2}{2}, u = \frac{2 - 2}{2}, u = - \frac{2 - 2}{2}

u = cos (x)

استبدل مجددًا

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

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

cos (x) = \frac{1}{2} : x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

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

Apply trig inverse properties

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

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

حلول عامّة لـ

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

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (x) = - \frac{1}{2} : x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

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

Apply trig inverse properties

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

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

حلول عامّة لـ

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

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

cos (x) = \frac{2 + 2}{2} : x = arccos (\frac{2 + 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 + 2}{2}) + 2 πn

cos (x) = \frac{2 + 2}{2}

Apply trig inverse properties

cos (x) = \frac{2 + 2}{2}

cos (x) = \frac{2 + 2}{2} :

حلول عامّة لـ

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

x = arccos (\frac{2 + 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 + 2}{2}) + 2 πn

x = arccos (\frac{2 + 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 + 2}{2}) + 2 πn

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

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

Apply trig inverse properties

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

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

حلول عامّة لـ

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

x = arccos (- \frac{2 + 2}{2}) + 2 πn, x = - arccos (- \frac{2 + 2}{2}) + 2 πn

x = arccos (- \frac{2 + 2}{2}) + 2 πn, x = - arccos (- \frac{2 + 2}{2}) + 2 πn

cos (x) = \frac{2 - 2}{2} : x = arccos (\frac{2 - 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 - 2}{2}) + 2 πn

cos (x) = \frac{2 - 2}{2}

Apply trig inverse properties

cos (x) = \frac{2 - 2}{2}

cos (x) = \frac{2 - 2}{2} :

حلول عامّة لـ

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

x = arccos (\frac{2 - 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 - 2}{2}) + 2 πn

x = arccos (\frac{2 - 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 - 2}{2}) + 2 πn

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

cos (x) = - \frac{2 - 2}{2}

Apply trig inverse properties

cos (x) = - \frac{2 - 2}{2}

cos (x) = - \frac{2 - 2}{2} :

حلول عامّة لـ

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

x = arccos (- \frac{2 - 2}{2}) + 2 πn, x = - arccos (- \frac{2 - 2}{2}) + 2 πn

x = arccos (- \frac{2 - 2}{2}) + 2 πn, x = - arccos (- \frac{2 - 2}{2}) + 2 πn

وحّد الحلول

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn, x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn, x = arccos (\frac{2 + 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 + 2}{2}) + 2 πn, x = arccos (- \frac{2 + 2}{2}) + 2 πn, x = - arccos (- \frac{2 + 2}{2}) + 2 πn, x = arccos (\frac{2 - 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 - 2}{2}) + 2 πn, x = arccos (- \frac{2 - 2}{2}) + 2 πn, x = - arccos (- \frac{2 - 2}{2}) + 2 πn

أظهر الحلّ بالتمثيل العشريّ

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn, x = 0.39269 \dots + 2 πn, x = 2 π - 0.39269 \dots + 2 πn, x = 2.74889 \dots + 2 πn, x = - 2.74889 \dots + 2 πn, x = 1.17809 \dots + 2 πn, x = 2 π - 1.17809 \dots + 2 πn, x = 1.96349 \dots + 2 πn, x = - 1.96349 \dots + 2 πn

رسم

أمثلة شائعة

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