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شائع علم المثلثات >

((1-cos(x))(csc^2(x)-4))/(1+cos(x))=0

الحلّ

\frac{( 1 - cos ( x ) ) ( csc ^{2} ( x ) - 4 )}{1 + cos ( x )} = 0

الحلّ

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

درجات

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

خطوات الحلّ

\frac{( 1 - cos ( x ) ) ( csc ^{2} ( x ) - 4 )}{1 + cos ( x )} = 0

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

(1 - cos (x)) (csc^{2} (x) - 4) = 0

حلّ كل جزء على حدة

1 - cos (x) = 0 or csc^{2} (x) - 4 = 0

1 - cos (x) = 0 : x = 2 πn

1 - cos (x) = 0

انقل

1

إلى الجانب الأيمن

1 - cos (x) = 0

من الطرفين

1

اطرح

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

بسّط

- cos (x) = - 1

- cos (x) = - 1

- 1

اقسم الطرفين على

- cos (x) = - 1

- 1

اقسم الطرفين على

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

بسّط

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 2 πn

x = 0 + 2 πn

حلّ:

x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

csc^{2} (x) - 4 = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

بالاستعانة بطريقة التعويض

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

csc (x) = u :

على افتراض أنّ

u^{2} - 4 = 0

u^{2} - 4 = 0 : u = 2, u = - 2

u^{2} - 4 = 0

انقل

4

إلى الجانب الأيمن

u^{2} - 4 = 0

للطرفين

4

أضف

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

بسّط

u^{2} = 4

u^{2} = 4

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

الحلول هي

x^{2} = f (a)

لـ

u = 4, u = - 4

4 = 2

4

4 = 2^{2} :

حلّل العدد لعوامله أوّليّة

= 2^{2}

n a^{n} = a

:فعْل قانون الجذور

2^{2} = 2

= 2

- 4 = - 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

= - 2

u = 2, u = - 2

u = csc (x)

استبدل مجددًا

csc (x) = 2, csc (x) = - 2

csc (x) = 2, csc (x) = - 2

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

csc (x) = - 2 : x = \frac{7 π}{6} + 2 πn, x = \frac{11 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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

وحّد الحلول

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

وحّد الحلول

x = 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

2 πn :

بما أنّ المعادلة غير معرّفة لـ

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

رسم

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