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شائع علم المثلثات >

(sin(x))^{(sin(x))}= 1/(sqrt(2))

الحلّ

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

الحلّ

x = 0.25268 \dots + 2 πn, x = π - 0.25268 \dots + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

درجات

x = 14.47751 \dots^{\circ} + 36 0^{\circ} n, x = 165.52248 \dots^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

خطوات الحلّ

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

من الطرفين

\frac{1}{2}

اطرح

sin^{s i n (x)} (x) - \frac{1}{2} = 0

sin^{s i n (x)} (x) - \frac{1}{2}

بسّط:

\frac{2 sin ^{s i n (x)} ( x ) - 1}{2}

sin^{s i n (x)} (x) - \frac{1}{2}

sin^{s i n (x)} (x) = \frac{sin ^{s i n (x)} ( x ) 2}{2}

:حوّل الأعداد لكسور

= \frac{sin ^{s i n (x)} ( x ) 2}{2} - \frac{1}{2}

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

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{sin ^{s i n (x)} ( x ) 2 - 1}{2}

\frac{2 sin ^{s i n (x)} ( x ) - 1}{2} = 0

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

2 sin^{s i n (x)} (x) - 1 = 0

Rewrite using trig identities

- 1 + sin^{s i n (x)} (x) 2

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

:Use the basic trigonometric identity

= - 1 + (\frac{1}{csc ( x )})^{\frac{1}{c s c ( x )}} 2

- 1 + (\frac{1}{csc ( x )})^{\frac{1}{c s c ( x )}} 2 = 0

بالاستعانة بطريقة التعويض

- 1 + (\frac{1}{csc ( x )})^{\frac{1}{c s c ( x )}} 2 = 0

csc (x) = u :

على افتراض أنّ

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

- 1 + (\frac{1}{u})^{\frac{1}{u}} 2 = 0 : u = 4, u = 2

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

فعّل قانون القوى

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

f (x)^{g (x)} = e^{g (x) l n (f (x))}

:فعّل قانون القوى

(\frac{1}{u})^{\frac{1}{u}} = e^{\frac{1}{u} l n (\frac{1}{u})}

- 1 + e^{\frac{1}{u} l n (\frac{1}{u})} 2 = 0

- 1 + e^{\frac{1}{u} l n (\frac{1}{u})} 2 = 0

للطرفين

1

أضف

- 1 + e^{\frac{1}{u} l n (\frac{1}{u})} 2 + 1 = 0 + 1

بسّط

2 e^{\frac{1}{u} l n (\frac{1}{u})} = 1

2

اقسم الطرفين على

2 e^{\frac{1}{u} l n (\frac{1}{u})} = 1

2

اقسم الطرفين على

\frac{2 e ^{\frac{1}{u} l n (\frac{1}{u})}}{2} = \frac{1}{2}

بسّط

e^{\frac{1}{u} l n (\frac{1}{u})} = \frac{1}{2}

e^{\frac{1}{u} l n (\frac{1}{u})} = \frac{1}{2}

بسّط

e^{\frac{1}{u} l n (\frac{1}{u})} = \frac{2}{2}

فعّل قانون القوى

e^{\frac{1}{u} l n (\frac{1}{u})} = \frac{2}{2}

Eq u a t i o n 1

للأساس

2

حوّل:

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

بسّط

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

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

ln (f (x)) = ln (g (x))

إذا

, f (x) = g (x)

إذا تحقّق أنّ

ln (e^{\frac{1}{u} l n (\frac{1}{u})}) = ln (2^{\frac{- 1}{2}})

ln (e^{a}) = a

:فعّل قانون اللوغارتمات

ln (e^{\frac{1}{u} l n (\frac{1}{u})}) = \frac{1}{u} ln (\frac{1}{u})

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

ln (x^{a}) = a \cdot ln (x)

:فعّل قانون اللوغارتمات

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

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

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

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

حلّ:

u = 4, u = 2

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

u

اضرب الطرفين بـ

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

u

اضرب الطرفين بـ

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

بسّط

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

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

بسّط:

ln (\frac{1}{u})

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

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

:اضرب كسور

= \frac{1 \cdot ln ( \frac{1}{u} ) u}{u}

u :

إلغ العوامل المشتركة

= 1 \cdot ln (\frac{1}{u})

1 \cdot ln (\frac{1}{u}) = ln (\frac{1}{u}) :

اضرب

= ln (\frac{1}{u})

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

بسّط:

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

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

\frac{- a}{b} = - \frac{a}{b}

: استخدم ميزات الكسور التالية

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

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

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

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

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

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

e^{l n (\frac{1}{u})}

بسّط:

\frac{1}{u}

e^{l n (\frac{1}{u})}

a^{l o g_{a} (b)} = b

:فعّل قانون اللوغارتمات

= \frac{1}{u}

e^{- \frac{1}{2} u l n (2)}

بسّط:

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

e^{- \frac{1}{2} u l n (2)}

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

:فعّل قانون القوى

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

a^{l o g_{a} (b)} = b

:فعّل قانون اللوغارتمات

e^{l n (2)} = 2

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

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

u

اضرب الطرفين بـ

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

بسّط

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

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

حلّ:

u = 4, u = 2

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

Prepare

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

for Lambert form:

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

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

x e^{x} = a

is equation in Lambert form

فعّل قانون القوى

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

Eq u a t i o n 1

للأساس

e

حوّل:

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

a = b^{l o g_{b} (a)}

:فعّل قانون القوى

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

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

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

:فعّل قانون القوى

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

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

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

بسّط

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

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

u = - \frac{2 v}{ln ( 2 )}

وكذلك

- \frac{1}{2} u ln (2) = v

اكتب المعادلة مجددًا، بحيث أنّ

1 = e^{v} (- \frac{2 v}{ln ( 2 )})

Rewrite

1 = e^{v} (- \frac{2 v}{ln ( 2 )})

in Lambert form:

e^{v} v = - \frac{ln ( 2 )}{2}

1 = e^{v} (- \frac{2 v}{ln ( 2 )})

x e^{x} = a

is equation in Lambert form

بدّل الأطراف

e^{v} (- \frac{2 v}{ln ( 2 )}) = 1

ln (2)

اضرب الطرفين بـ

e^{v} (- \frac{2 v}{ln ( 2 )}) ln (2) = 1 \cdot ln (2)

بسّط

- 2 e^{v} v = ln (2)

- 2

اقسم الطرفين على

\frac{- 2 e ^{v} v}{- 2} = \frac{ln ( 2 )}{- 2}

بسّط

e^{v} v = - \frac{ln ( 2 )}{2}

e^{v} v = - \frac{ln ( 2 )}{2}

حلّ:

v = - 2 ln (2), v = - ln (2)

e^{v} v = - \frac{ln ( 2 )}{2}

Solutions for

x e^{x} = a

where

- \frac{1}{e} \leq a < 0

are principal and negative branches of Lambert

W

function:

x = W_{0} (a), W_{- 1} (a)

v = W_{- 1} (- \frac{ln ( 2 )}{2}), v = W_{0} (- \frac{ln ( 2 )}{2})

بسّط

v = - 2 ln (2), v = - ln (2)

v = - 2 ln (2), v = - ln (2)

Substitute back

v = - \frac{1}{2} u ln (2),

solve for

u

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

حلّ:

u = 4

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

- 2

اضرب الطرفين بـ

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

- 2

اضرب الطرفين بـ

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

بسّط

ln (2) u = 4 ln (2)

ln (2) u = 4 ln (2)

ln (2)

اقسم الطرفين على

ln (2) u = 4 ln (2)

ln (2)

اقسم الطرفين على

\frac{ln ( 2 ) u}{ln ( 2 )} = \frac{4 ln ( 2 )}{ln ( 2 )}

بسّط

u = 4

u = 4

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

حلّ:

u = 2

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

- 2

اضرب الطرفين بـ

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

- 2

اضرب الطرفين بـ

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

بسّط

ln (2) u = 2 ln (2)

ln (2) u = 2 ln (2)

ln (2)

اقسم الطرفين على

ln (2) u = 2 ln (2)

ln (2)

اقسم الطرفين على

\frac{ln ( 2 ) u}{ln ( 2 )} = \frac{2 ln ( 2 )}{ln ( 2 )}

بسّط

u = 2

u = 2

u = 4, u = 2

u = 4, u = 2

افحص الإجبات:

u = 4

صحيح

, u = 2

صحيح

للتحقّق من دقّة الحلول

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

عوّض الحلول في

إلغي الحلول التي تعطي قضيّة كذب

u = 4

استبدل:

صحيح

\frac{1}{4} ln (\frac{1}{4}) = \frac{- 1}{2} ln (2)

\frac{1}{4} ln (\frac{1}{4}) = - \frac{1}{2} ln (2)

\frac{1}{4} ln (\frac{1}{4})

ln (\frac{1}{4})

بسّط:

- 2 ln (2)

ln (\frac{1}{4})

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

:فعّل قانون اللوغارتمات

= - ln (4)

4 = 2^{2}

بصيغة أساس وقوى

4

اكتب

= - ln (2^{2})

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

:فعّل قانون اللوغارتمات

ln (2^{2}) = 2 ln (2)

= - 2 ln (2)

= \frac{1}{4} (- 2 ln (2))

(- a) = - a

:احذف الأقواس

= - \frac{1}{4} \cdot 2 ln (2)

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

:اضرب كسور

= - \frac{1 \cdot 2}{4} ln (2)

\frac{1 \cdot 2}{4} = \frac{1}{2}

\frac{1 \cdot 2}{4}

1 \cdot 2 = 2 :

اضرب الأعداد

= \frac{2}{4}

2 :

إلغ العوامل المشتركة

= \frac{1}{2}

= - \frac{1}{2} ln (2)

\frac{- 1}{2} ln (2) = - \frac{1}{2} ln (2)

\frac{- 1}{2} ln (2)

\frac{- a}{b} = - \frac{a}{b}

: استخدم ميزات الكسور التالية

= - \frac{1}{2} ln (2)

- \frac{1}{2} ln (2) = - \frac{1}{2} ln (2)

صحيح

u = 2

استبدل:

صحيح

\frac{1}{2} ln (\frac{1}{2}) = \frac{- 1}{2} ln (2)

\frac{1}{2} ln (\frac{1}{2}) = - \frac{1}{2} ln (2)

\frac{1}{2} ln (\frac{1}{2})

ln (\frac{1}{2})

بسّط:

- ln (2)

ln (\frac{1}{2})

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

:فعّل قانون اللوغارتمات

= - ln (2)

= \frac{1}{2} (- ln (2))

(- a) = - a

:احذف الأقواس

= - \frac{1}{2} ln (2)

\frac{- 1}{2} ln (2) = - \frac{1}{2} ln (2)

\frac{- 1}{2} ln (2)

\frac{- a}{b} = - \frac{a}{b}

: استخدم ميزات الكسور التالية

= - \frac{1}{2} ln (2)

- \frac{1}{2} ln (2) = - \frac{1}{2} ln (2)

صحيح

The solutions are

u = 4, u = 2

u = 4, u = 2

u = csc (x)

استبدل مجددًا

csc (x) = 4, csc (x) = 2

csc (x) = 4, csc (x) = 2

csc (x) = 4 : x = arccsc (4) + 2 πn, x = π - arccsc (4) + 2 πn

csc (x) = 4

Apply trig inverse properties

csc (x) = 4

csc (x) = 4 :

حلول عامّة لـ

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

x = arccsc (4) + 2 πn, x = π - arccsc (4) + 2 πn

x = arccsc (4) + 2 πn, x = π - arccsc (4) + 2 πn

csc (x) = 2 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

csc (x) = 2

csc (x) = 2 :

حلول عامّة لـ

csc (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

وحّد الحلول

x = arccsc (4) + 2 πn, x = π - arccsc (4) + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

أظهر الحلّ بالتمثيل العشريّ

x = 0.25268 \dots + 2 πn, x = π - 0.25268 \dots + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

رسم

أمثلة شائعة

(d^2-4)=cos^2(x)

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

cos(3θ)=4cos(3θ)-3cos(θ)

cos (3 θ) = 4 cos (3 θ) - 3 cos (θ)

sin^2(x)-7sin(x)=0

sin^{2} (x) - 7 sin (x) = 0

sin(x)=(48sin(69))/(47.5)

sin (x) = \frac{48 sin ( 6 9 ^{\circ} )}{47.5}

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

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