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شائع علم المثلثات >

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

الحلّ

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

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

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

من الطرفين

3

اطرح

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

a^{b} = a^{2} a^{b - 2} :

فعّل قانون القوى

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

Rewrite using trig identities

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

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

:فعّل نطريّة فيتاغوروس

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

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

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

بسّط:

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

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

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

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

a^{b} \cdot a^{c} = a^{b + c}

:فعّل قانون القوى

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

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

1 + 1 = 2 :

اجمع الأعداد

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

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

(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)

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

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

وسٌع:

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

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

فعّل قانون ضرب الأقواس

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

فعّل قوانين سالب-موجب

+ (- a) = - a

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

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

بسّط:

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

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

2 \cdot 1 = 2 :

اضرب الأعداد

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

2 \cdot 2 = 4 :

اضرب الأعداد

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

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

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

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

بسّط:

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

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

جمّع التعابير المتشابهة

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

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

اجمع العناصر المتشابهة

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

- 1 + 2 = 1 :

اطرح/اجمع الأعداد

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

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

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

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

بالاستعانة بطريقة التعويض

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

sin (x) = u :

على افتراض أنّ

1 - 4 u^{2} = 0

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

1 - 4 u^{2} = 0

انقل

1

إلى الجانب الأيمن

1 - 4 u^{2} = 0

من الطرفين

1

اطرح

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

بسّط

- 4 u^{2} = - 1

- 4 u^{2} = - 1

- 4

اقسم الطرفين على

- 4 u^{2} = - 1

- 4

اقسم الطرفين على

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

بسّط

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

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}

u = sin (x)

استبدل مجددًا

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

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

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}{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

csc((2x)/3)-1=0

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

6tan(x)=tan^2(x),x=0

6 tan (x) = tan^{2} (x), x = 0

5cos^2(x)+3cos(x)-1=0

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

3cot(x)+5=8

3 cot (x) + 5 = 8

solvefor a,sin(arccos(x))=sqrt(1-x^2)

so l v e f or a, sin (arccos (x)) = 1 - x^{2}