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شائع علم المثلثات >

csc(x-pi)cos^2(x-pi)-cot(x-pi)=0

الحلّ

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

الحلّ

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

درجات

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

خطوات الحلّ

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

Rewrite using trig identities

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

Rewrite using trig identities

cos (x - π)

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

:فعّل متطابقة الطرح لزوايا

= cos (x) cos (π) + sin (x) sin (π)

cos (x) cos (π) + sin (x) sin (π)

بسّط:

- cos (x)

cos (x) cos (π) + sin (x) sin (π)

cos (x) cos (π) = - cos (x)

cos (x) cos (π)

cos (π)

بسّط:

- 1

cos (π)

استخدم المتطابقة الأساسيّة التالية:

cos (π) = (- 1)

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - 1

= (- 1) cos (x)

بسّط

= - cos (x)

= - cos (x) + sin (π) sin (x)

sin (x) sin (π) = 0

sin (x) sin (π)

sin (π)

بسّط:

0

sin (π)

استخدم المتطابقة الأساسيّة التالية:

sin (π) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0 \cdot sin (x)

0 \cdot a = 0

فعّل القانون

= 0

= - cos (x) + 0

- cos (x) + 0 = - cos (x)

= - cos (x)

= - cos (x)

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

:Use the basic trigonometric identity

= \frac{1}{sin ( x - π )}

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

:فعّل متطابقة الطرح لزوايا

= \frac{1}{sin ( x ) cos ( π ) - cos ( x ) sin ( π )}

\frac{1}{sin ( x ) cos ( π ) - cos ( x ) sin ( π )}

بسّط:

- \frac{1}{sin ( x )}

\frac{1}{sin ( x ) cos ( π ) - cos ( x ) sin ( π )}

sin (x) cos (π) - cos (x) sin (π) = - sin (x)

sin (x) cos (π) - cos (x) sin (π)

sin (x) cos (π) = - sin (x)

sin (x) cos (π)

cos (π)

بسّط:

- 1

cos (π)

استخدم المتطابقة الأساسيّة التالية:

cos (π) = (- 1)

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - 1

= (- 1) sin (x)

بسّط

= - sin (x)

= - sin (x) - sin (π) cos (x)

cos (x) sin (π) = 0

cos (x) sin (π)

sin (π)

بسّط:

0

sin (π)

استخدم المتطابقة الأساسيّة التالية:

sin (π) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0 \cdot cos (x)

0 \cdot a = 0

فعّل القانون

= 0

= - sin (x) - 0

- sin (x) - 0 = - sin (x)

= - sin (x)

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

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

: استخدم ميزات الكسور التالية

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

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

cot (x) = \frac{cos ( x )}{sin ( x )}

:Use the basic trigonometric identity

= \frac{cos ( x - π )}{sin ( x - π )}

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

:فعّل متطابقة الطرح لزوايا

= \frac{cos ( x - π )}{sin ( x ) cos ( π ) - cos ( x ) sin ( π )}

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

:فعّل متطابقة الطرح لزوايا

= \frac{cos ( x ) cos ( π ) + sin ( x ) sin ( π )}{sin ( x ) cos ( π ) - cos ( x ) sin ( π )}

\frac{cos ( x ) cos ( π ) + sin ( x ) sin ( π )}{sin ( x ) cos ( π ) - cos ( x ) sin ( π )}

بسّط:

\frac{cos ( x )}{sin ( x )}

\frac{cos ( x ) cos ( π ) + sin ( x ) sin ( π )}{sin ( x ) cos ( π ) - cos ( x ) sin ( π )}

cos (x) cos (π) + sin (x) sin (π) = - cos (x)

cos (x) cos (π) + sin (x) sin (π)

cos (x) cos (π) = - cos (x)

cos (x) cos (π)

cos (π)

بسّط:

- 1

cos (π)

استخدم المتطابقة الأساسيّة التالية:

cos (π) = (- 1)

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - 1

= (- 1) cos (x)

بسّط

= - cos (x)

= - cos (x) + sin (π) sin (x)

sin (x) sin (π) = 0

sin (x) sin (π)

sin (π)

بسّط:

0

sin (π)

استخدم المتطابقة الأساسيّة التالية:

sin (π) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0 \cdot sin (x)

0 \cdot a = 0

فعّل القانون

= 0

= - cos (x) + 0

- cos (x) + 0 = - cos (x)

= - cos (x)

= \frac{- cos ( x )}{cos ( π ) sin ( x ) - sin ( π ) cos ( x )}

sin (x) cos (π) - cos (x) sin (π) = - sin (x)

sin (x) cos (π) - cos (x) sin (π)

sin (x) cos (π) = - sin (x)

sin (x) cos (π)

cos (π)

بسّط:

- 1

cos (π)

استخدم المتطابقة الأساسيّة التالية:

cos (π) = (- 1)

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - 1

= (- 1) sin (x)

بسّط

= - sin (x)

= - sin (x) - sin (π) cos (x)

cos (x) sin (π) = 0

cos (x) sin (π)

sin (π)

بسّط:

0

sin (π)

استخدم المتطابقة الأساسيّة التالية:

sin (π) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0 \cdot cos (x)

0 \cdot a = 0

فعّل القانون

= 0

= - sin (x) - 0

- sin (x) - 0 = - sin (x)

= - sin (x)

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

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

: استخدم ميزات الكسور التالية

= \frac{cos ( x )}{sin ( x )}

= \frac{cos ( x )}{sin ( x )}

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

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

بسّط:

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

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

(- a) = - a

:احذف الأقواس

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

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

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

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

(- cos (x))^{2}

زوجيّ n

إذا تحقّق أنّ

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

:فعّل قانون القوى

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

= cos^{2} (x)

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

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

:اضرب كسور

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

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

اضرب

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

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

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

فعّل القانون

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

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

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

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

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

بالاستعانة بطريقة التعويض

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

cos (x) = u :

على افتراض أنّ

- u^{2} - u = 0

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

- u^{2} - u = 0

حلّ بالاستعانة بالصيغة التربيعيّة

- u^{2} - u = 0

الصيغة لحلّ المعادلة التربيعيّة

a = - 1, b = - 1, c = 0

لـ

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

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

(- 1)^{2} - 4 (- 1) \cdot 0

- (- a) = a

فعّل القانون

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

(- 1)^{2} = 1

(- 1)^{2}

زوجيّ n

إذا تحقّق أنّ

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

:فعّل قانون القوى

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

= 1^{2}

1^{a} = 1

فعّل القانون

= 1

4 \cdot 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

0 \cdot a = 0

فعّل القانون

= 0

= 1 + 0

1 + 0 = 1 :

اجمع الأعداد

= 1

1 = 1

فعّل القانون

= 1

u_{1, 2} = \frac{- ( - 1 ) \pm 1}{2 ( - 1 )}

Separate the solutions

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

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

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

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

:احذف الأقواس

= \frac{1 + 1}{- 2 \cdot 1}

1 + 1 = 2 :

اجمع الأعداد

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

2 \cdot 1 = 2 :

اضرب الأعداد

= \frac{2}{- 2}

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

: استخدم ميزات الكسور التالية

= - \frac{2}{2}

\frac{a}{a} = 1

فعّل القانون

= - 1

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

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

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

:احذف الأقواس

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

1 - 1 = 0 :

اطرح الأعداد

= \frac{0}{- 2 \cdot 1}

2 \cdot 1 = 2 :

اضرب الأعداد

= \frac{0}{- 2}

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

: استخدم ميزات الكسور التالية

= - \frac{0}{2}

\frac{0}{a} = 0, a \neq = 0

فعّل القانون

= - 0

= 0

حلول المعادلة التربيعيّة هي

u = - 1, u = 0

u = cos (x)

استبدل مجددًا

cos (x) = - 1, cos (x) = 0

cos (x) = - 1, cos (x) = 0

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

cos (x) = - 1 :

حلول عامّة لـ

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0 :

حلول عامّة لـ

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

وحّد الحلول

x = π + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

π + 2 πn :

بما أنّ المعادلة غير معرّفة لـ

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

رسم

أمثلة شائعة

tan^2(θ)+6tan(θ)+8=0

tan^{2} (θ) + 6 tan (θ) + 8 = 0

sin(x/2)+cos(x)=1

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

-cos(x)+cos(2x)=0

- cos (x) + cos (2 x) = 0

sin(4x)+1=0

sin (4 x) + 1 = 0

4cos^2(θ)=3

4 cos^{2} (θ) = 3