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شائع >

-3+4(cos^2(x))^2+3sin^2(x)=0

الحلّ

- 3 + 4 (cos^{2} (x))^{2} + 3 sin^{2} (x) = 0

الحلّ

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

درجات

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

- 3 + 4 (cos^{2} (x))^{2} + 3 sin^{2} (x) = 0

Rewrite using trig identities

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

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

:فعّل نطريّة فيتاغوروس

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

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

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

بسّط:

4 sin^{4} (x) - 5 sin^{2} (x) + 1

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

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

(1 - sin^{2} (x))^{2} : 1 - 2 sin^{2} (x) + sin^{4} (x)

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

:فعّل صيغة الضرب المختصر

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

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

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

بسّط:

1 - 2 sin^{2} (x) + sin^{4} (x)

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

1^{a} = 1

فعّل القانون

1^{2} = 1

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

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

2 \cdot 1 \cdot sin^{2} (x)

2 \cdot 1 = 2 :

اضرب الأعداد

= 2 sin^{2} (x)

(sin^{2} (x))^{2} = sin^{4} (x)

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

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

:فعّل قانون القوى

= sin^{2 \cdot 2} (x)

2 \cdot 2 = 4 :

اضرب الأعداد

= sin^{4} (x)

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

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

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

4 (1 - 2 sin^{2} (x) + sin^{4} (x))

وسٌع:

4 - 8 sin^{2} (x) + 4 sin^{4} (x)

4 (1 - 2 sin^{2} (x) + sin^{4} (x))

فعّل قانون ضرب الأقواس

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

فعّل قوانين سالب-موجب

+ (- a) = - a

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

4 \cdot 1 - 4 \cdot 2 sin^{2} (x) + 4 sin^{4} (x)

بسّط:

4 - 8 sin^{2} (x) + 4 sin^{4} (x)

4 \cdot 1 - 4 \cdot 2 sin^{2} (x) + 4 sin^{4} (x)

4 \cdot 1 = 4 :

اضرب الأعداد

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

4 \cdot 2 = 8 :

اضرب الأعداد

= 4 - 8 sin^{2} (x) + 4 sin^{4} (x)

= 4 - 8 sin^{2} (x) + 4 sin^{4} (x)

= - 3 + 4 - 8 sin^{2} (x) + 4 sin^{4} (x) + 3 sin^{2} (x)

- 3 + 4 - 8 sin^{2} (x) + 4 sin^{4} (x) + 3 sin^{2} (x)

بسّط:

4 sin^{4} (x) - 5 sin^{2} (x) + 1

- 3 + 4 - 8 sin^{2} (x) + 4 sin^{4} (x) + 3 sin^{2} (x)

جمّع التعابير المتشابهة

= - 8 sin^{2} (x) + 4 sin^{4} (x) + 3 sin^{2} (x) - 3 + 4

- 8 sin^{2} (x) + 3 sin^{2} (x) = - 5 sin^{2} (x) :

اجمع العناصر المتشابهة

= - 5 sin^{2} (x) + 4 sin^{4} (x) - 3 + 4

- 3 + 4 = 1 :

اطرح/اجمع الأعداد

= 4 sin^{4} (x) - 5 sin^{2} (x) + 1

= 4 sin^{4} (x) - 5 sin^{2} (x) + 1

= 4 sin^{4} (x) - 5 sin^{2} (x) + 1

1 + 4 sin^{4} (x) - 5 sin^{2} (x) = 0

بالاستعانة بطريقة التعويض

1 + 4 sin^{4} (x) - 5 sin^{2} (x) = 0

sin (x) = u :

على افتراض أنّ

1 + 4 u^{4} - 5 u^{2} = 0

1 + 4 u^{4} - 5 u^{2} = 0 : u = 1, u = - 1, u = \frac{1}{2}, u = - \frac{1}{2}

1 + 4 u^{4} - 5 u^{2} = 0

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

اكتب بالصورة الاعتياديّة

4 u^{4} - 5 u^{2} + 1 = 0

v^{2} = u^{4}

وكذلك

v = u^{2}

اكتب المعادلة مجددًا، بحيث أنّ

4 v^{2} - 5 v + 1 = 0

4 v^{2} - 5 v + 1 = 0

حلّ:

v = 1, v = \frac{1}{4}

4 v^{2} - 5 v + 1 = 0

حلّ بالاستعانة بالصيغة التربيعيّة

4 v^{2} - 5 v + 1 = 0

الصيغة لحلّ المعادلة التربيعيّة

a = 4, b = - 5, c = 1

لـ

v_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

v_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

(- 5)^{2} - 4 \cdot 4 \cdot 1 = 3

(- 5)^{2} - 4 \cdot 4 \cdot 1

زوجيّ n

إذا تحقّق أنّ

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

:فعّل قانون القوى

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 4 \cdot 1

4 \cdot 4 \cdot 1 = 16 :

اضرب الأعداد

= 5^{2} - 16

5^{2} = 25

= 25 - 16

25 - 16 = 9 :

اطرح الأعداد

= 9

9 = 3^{2} :

حلّل العدد لعوامله أوّليّة

= 3^{2}

n a^{n} = a

:فعْل قانون الجذور

3^{2} = 3

= 3

v_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 4}

Separate the solutions

v_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 4}, v_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 4}

v = \frac{- ( - 5 ) + 3}{2 \cdot 4} : 1

\frac{- ( - 5 ) + 3}{2 \cdot 4}

- (- a) = a

فعّل القانون

= \frac{5 + 3}{2 \cdot 4}

5 + 3 = 8 :

اجمع الأعداد

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

2 \cdot 4 = 8 :

اضرب الأعداد

= \frac{8}{8}

\frac{a}{a} = 1

فعّل القانون

= 1

v = \frac{- ( - 5 ) - 3}{2 \cdot 4} : \frac{1}{4}

\frac{- ( - 5 ) - 3}{2 \cdot 4}

- (- a) = a

فعّل القانون

= \frac{5 - 3}{2 \cdot 4}

5 - 3 = 2 :

اطرح الأعداد

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

2 \cdot 4 = 8 :

اضرب الأعداد

= \frac{2}{8}

2 :

إلغ العوامل المشتركة

= \frac{1}{4}

حلول المعادلة التربيعيّة هي

v = 1, v = \frac{1}{4}

v = 1, v = \frac{1}{4}

Substitute back

v = u^{2},

solve for

u

u^{2} = 1

حلّ:

u = 1, u = - 1

u^{2} = 1

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

الحلول هي

x^{2} = f (a)

لـ

u = 1, u = - 1

1 = 1

1

1 = 1

فعّل القانون

= 1

- 1 = - 1

- 1

1 = 1

فعّل القانون

= - 1

u = 1, u = - 1

u^{2} = \frac{1}{4}

حلّ:

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

u^{2} = \frac{1}{4}

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

الحلول هي

x^{2} = f (a)

لـ

u = \frac{1}{4}, u = - \frac{1}{4}

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

\frac{1}{4}

a \geq 0, b \geq 0

بافتراض أنّ

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

:فعّل قانون الجذور

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

1 = 1

فعّل القانون

= \frac{1}{2}

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

- \frac{1}{4}

\frac{1}{4}

بسّط:

\frac{1}{2}

\frac{1}{4}

a \geq 0, b \geq 0

بافتراض أنّ

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

:فعّل قانون الجذور

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

= - \frac{1}{2}

1 = 1

فعّل القانون

= - \frac{1}{2}

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

The solutions are

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

u = sin (x)

استبدل مجددًا

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

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

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

sin (x) = 1 :

حلول عامّة لـ

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}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

sin (x) = - 1 :

حلول عامّة لـ

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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

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

حلول عامّة لـ

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}

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

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

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

حلول عامّة لـ

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}

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}{2} + 2 πn, x = \frac{3 π}{2} + 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

رسم

أمثلة شائعة

4sin(x-30)-1.3=0

4 sin (x - 3 0^{\circ}) - 1.3 = 0

2cos(θ/2)=cos(θ)+1

2 cos (\frac{θ}{2}) = cos (θ) + 1

cos(sqrt(3)x)+sin(sqrt(3)x)=0

cos (3 x) + sin (3 x) = 0

1/(sqrt(3))=tan(x)

\frac{1}{3} = tan (x)

sin(θ)=-(sqrt(5))/5

sin (θ) = - \frac{5}{5}