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شائع علم المثلثات >

d=(sin^4(x)-cos^2(x)+5)/(4*cos^2(x))

الحلّ

d = \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 \cdot cos ^{2} ( x )}

الحلّ

x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

خطوات الحلّ

d = \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )}

بدّل الأطراف

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} = d

من الطرفين

d

اطرح

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} - d = 0

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} - d

بسّط:

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5 - 4 d cos ^{2} ( x )}{4 cos ^{2} ( x )}

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} - d

d = \frac{d 4 cos ^{2} ( x )}{4 cos ^{2} ( x )}

:حوّل الأعداد لكسور

= \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} - \frac{d \cdot 4 cos ^{2} ( x )}{4 cos ^{2} ( x )}

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

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5 - d \cdot 4 cos ^{2} ( x )}{4 cos ^{2} ( x )}

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5 - 4 d cos ^{2} ( x )}{4 cos ^{2} ( x )} = 0

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

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

Rewrite using trig identities

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

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

:فعّل نطريّة فيتاغوروس

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

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

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

بسّط:

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

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

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

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

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

افتح أقواس

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

فعّل قوانين سالب-موجب

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

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

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

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

وسٌع:

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

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

a (b - c) = ab - a c

: افتح أقواس بالاستعانة بـ

a = - 4 d, b = 1, c = sin^{2} (x)

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

فعّل قوانين سالب-موجب

- (- a) = a

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

4 \cdot 1 = 4 :

اضرب الأعداد

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

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

5 - 1 = 4 :

اطرح الأعداد

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

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

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

بالاستعانة بطريقة التعويض

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

sin (x) = u :

على افتراض أنّ

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

4 + u^{2} + u^{4} - 4 d + 4 u^{2} d = 0 : u = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

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

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

اكتب بالصورة الاعتياديّة

u^{4} + (1 + 4 d) u^{2} + 4 - 4 d = 0

v^{2} = u^{4}

وكذلك

v = u^{2}

اكتب المعادلة مجددًا، بحيث أنّ

v^{2} + (1 + 4 d) v + 4 - 4 d = 0

v^{2} + (1 + 4 d) v + 4 - 4 d = 0

حلّ:

v = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, v = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

v^{2} + (1 + 4 d) v + 4 - 4 d = 0

حلّ بالاستعانة بالصيغة التربيعيّة

v^{2} + (1 + 4 d) v + 4 - 4 d = 0

الصيغة لحلّ المعادلة التربيعيّة

a = 1, b = 1 + 4 d, c = 4 - 4 d

لـ

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

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

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

بسّط:

16 d^{2} + 24 d - 15

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

4 \cdot 1 = 4 :

اضرب الأعداد

= (4 d + 1)^{2} - 4 (- 4 d + 4)

(1 + 4 d)^{2} - 4 (4 - 4 d)

وسٌع:

16 d^{2} + 24 d - 15

(1 + 4 d)^{2} - 4 (4 - 4 d)

(1 + 4 d)^{2} : 1 + 8 d + 16 d^{2}

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

:فعّل صيغة الضرب المختصر

a = 1, b = 4 d

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

1^{2} + 2 \cdot 1 \cdot 4 d + (4 d)^{2}

بسّط:

1 + 8 d + 16 d^{2}

1^{2} + 2 \cdot 1 \cdot 4 d + (4 d)^{2}

1^{a} = 1

فعّل القانون

1^{2} = 1

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

2 \cdot 1 \cdot 4 d = 8 d

2 \cdot 1 \cdot 4 d

2 \cdot 1 \cdot 4 = 8 :

اضرب الأعداد

= 8 d

(4 d)^{2} = 16 d^{2}

(4 d)^{2}

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

:فعّل قانون القوى

= 4^{2} d^{2}

4^{2} = 16

= 16 d^{2}

= 1 + 8 d + 16 d^{2}

= 1 + 8 d + 16 d^{2}

= 1 + 8 d + 16 d^{2} - 4 (4 - 4 d)

- 4 (4 - 4 d)

وسٌع:

- 16 + 16 d

- 4 (4 - 4 d)

a (b - c) = ab - a c

: افتح أقواس بالاستعانة بـ

a = - 4, b = 4, c = 4 d

= - 4 \cdot 4 - (- 4) \cdot 4 d

فعّل قوانين سالب-موجب

- (- a) = a

= - 4 \cdot 4 + 4 \cdot 4 d

4 \cdot 4 = 16 :

اضرب الأعداد

= - 16 + 16 d

= 1 + 8 d + 16 d^{2} - 16 + 16 d

1 + 8 d + 16 d^{2} - 16 + 16 d

بسّط:

16 d^{2} + 24 d - 15

1 + 8 d + 16 d^{2} - 16 + 16 d

جمّع التعابير المتشابهة

= 16 d^{2} + 8 d + 16 d + 1 - 16

8 d + 16 d = 24 d :

اجمع العناصر المتشابهة

= 16 d^{2} + 24 d + 1 - 16

1 - 16 = - 15 :

اطرح/اجمع الأعداد

= 16 d^{2} + 24 d - 15

= 16 d^{2} + 24 d - 15

= 16 d^{2} + 24 d - 15

v_{1, 2} = \frac{- ( 1 + 4 d ) \pm 16 d ^{2} + 24 d - 15}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( 1 + 4 d ) + 16 d ^{2} + 24 d - 15}{2 \cdot 1}, v_{2} = \frac{- ( 1 + 4 d ) - 16 d ^{2} + 24 d - 15}{2 \cdot 1}

v = \frac{- ( 1 + 4 d ) + 16 d ^{2} + 24 d - 15}{2 \cdot 1} : \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

\frac{- ( 1 + 4 d ) + 16 d ^{2} + 24 d - 15}{2 \cdot 1}

2 \cdot 1 = 2 :

اضرب الأعداد

= \frac{- ( 4 d + 1 ) + 16 d ^{2} + 24 d - 15}{2}

- (1 + 4 d) : - 1 - 4 d

- (1 + 4 d)

افتح أقواس

= - (1) - (4 d)

فعّل قوانين سالب-موجب

+ (- a) = - a

= - 1 - 4 d

= \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

v = \frac{- ( 1 + 4 d ) - 16 d ^{2} + 24 d - 15}{2 \cdot 1} : \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

\frac{- ( 1 + 4 d ) - 16 d ^{2} + 24 d - 15}{2 \cdot 1}

2 \cdot 1 = 2 :

اضرب الأعداد

= \frac{- ( 4 d + 1 ) - 16 d ^{2} + 24 d - 15}{2}

- (1 + 4 d) : - 1 - 4 d

- (1 + 4 d)

افتح أقواس

= - (1) - (4 d)

فعّل قوانين سالب-موجب

+ (- a) = - a

= - 1 - 4 d

= \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

حلول المعادلة التربيعيّة هي

v = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, v = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

v = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, v = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

Substitute back

v = u^{2},

solve for

u

u^{2} = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

حلّ:

u = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

u^{2} = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

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

الحلول هي

x^{2} = f (a)

لـ

u = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

u^{2} = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

حلّ:

u = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

u^{2} = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

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

الحلول هي

x^{2} = f (a)

لـ

u = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

The solutions are

u = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

u = sin (x)

استبدل مجددًا

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} : x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

Apply trig inverse properties

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} :

حلول عامّة لـ

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} : x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

Apply trig inverse properties

sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} :

حلول عامّة لـ

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} : x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

Apply trig inverse properties

sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} :

حلول عامّة لـ

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} : x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

Apply trig inverse properties

sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} :

حلول عامّة لـ

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

وحّد الحلول

x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

رسم

أمثلة شائعة

2+cos^2(x)=5sin(x)

2 + cos^{2} (x) = 5 sin (x)

tan^3(3x)-2sin^3(3x)=0

tan^{3} (3 x) - 2 sin^{3} (3 x) = 0

cot^5(x)=(-1)/((sqrt(3)))

cot^{5} (x) = \frac{- 1}{( 3 )}

2cos^4(x)cos(x)-cos^5(x)=1

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

cos^4(x)-2sin^2(x)-1=0

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