English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

شائع علم المثلثات >

4cos^2(x)+sqrt(3)=2(sqrt(3)+1)

الحلّ

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

الحلّ

x = 0.26179 \dots + 2 πn, x = 2 π - 0.26179 \dots + 2 πn, x = 2.87979 \dots + 2 πn, x = - 2.87979 \dots + 2 πn

درجات

x = 1 5^{\circ} + 36 0^{\circ} n, x = 34 5^{\circ} + 36 0^{\circ} n, x = 16 5^{\circ} + 36 0^{\circ} n, x = - 16 5^{\circ} + 36 0^{\circ} n

خطوات الحلّ

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

بالاستعانة بطريقة التعويض

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

cos (x) = u :

على افتراض أنّ

4 u^{2} + 3 = 2 (3 + 1)

4 u^{2} + 3 = 2 (3 + 1) : u = \frac{3 + 2}{2}, u = - \frac{3 + 2}{2}

4 u^{2} + 3 = 2 (3 + 1)

انقل

3

إلى الجانب الأيمن

4 u^{2} + 3 = 2 (3 + 1)

من الطرفين

3

اطرح

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

بسّط

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

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

4

اقسم الطرفين على

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

4

اقسم الطرفين على

\frac{4 u ^{2}}{4} = \frac{2 ( 3 + 1 )}{4} - \frac{3}{4}

بسّط

\frac{4 u ^{2}}{4} = \frac{2 ( 3 + 1 )}{4} - \frac{3}{4}

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

بسّط:

u^{2}

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

\frac{4}{4} = 1 :

اقسم الأعداد

= u^{2}

\frac{2 ( 3 + 1 )}{4} - \frac{3}{4}

بسّط:

\frac{3 + 2}{4}

\frac{2 ( 3 + 1 )}{4} - \frac{3}{4}

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

فعّل القانون

= \frac{2 ( 1 + 3 ) - 3}{4}

2 (3 + 1) - 3

وسٌع:

3 + 2

2 (3 + 1) - 3

2 (3 + 1)

وسٌع:

23 + 2

2 (3 + 1)

a (b + c) = ab + a c

: افتح أقواس بالاستعانة بـ

a = 2, b = 3, c = 1

= 23 + 2 \cdot 1

2 \cdot 1 = 2 :

اضرب الأعداد

= 23 + 2

= 23 + 2 - 3

23 - 3 = 3 :

اجمع العناصر المتشابهة

= 3 + 2

= \frac{3 + 2}{4}

u^{2} = \frac{3 + 2}{4}

u^{2} = \frac{3 + 2}{4}

u^{2} = \frac{3 + 2}{4}

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

الحلول هي

x^{2} = f (a)

لـ

u = \frac{3 + 2}{4}, u = - \frac{3 + 2}{4}

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

\frac{3 + 2}{4}

a \geq 0, b \geq 0

بافتراض أنّ

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

:فعّل قانون الجذور

= \frac{3 + 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

= \frac{3 + 2}{2}

- \frac{3 + 2}{4} = - \frac{3 + 2}{2}

- \frac{3 + 2}{4}

\frac{3 + 2}{4}

بسّط:

\frac{3 + 2}{2}

\frac{3 + 2}{4}

a \geq 0, b \geq 0

بافتراض أنّ

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

:فعّل قانون الجذور

= \frac{3 + 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

= \frac{3 + 2}{2}

= - \frac{2 + 3}{2}

= - \frac{3 + 2}{2}

u = \frac{3 + 2}{2}, u = - \frac{3 + 2}{2}

u = cos (x)

استبدل مجددًا

cos (x) = \frac{3 + 2}{2}, cos (x) = - \frac{3 + 2}{2}

cos (x) = \frac{3 + 2}{2}, cos (x) = - \frac{3 + 2}{2}

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

cos (x) = \frac{3 + 2}{2}

Apply trig inverse properties

cos (x) = \frac{3 + 2}{2}

cos (x) = \frac{3 + 2}{2} :

حلول عامّة لـ

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{3 + 2}{2}) + 2 πn, x = 2 π - arccos (\frac{3 + 2}{2}) + 2 πn

x = arccos (\frac{3 + 2}{2}) + 2 πn, x = 2 π - arccos (\frac{3 + 2}{2}) + 2 πn

cos (x) = - \frac{3 + 2}{2} : x = arccos (- \frac{3 + 2}{2}) + 2 πn, x = - arccos (- \frac{3 + 2}{2}) + 2 πn

cos (x) = - \frac{3 + 2}{2}

Apply trig inverse properties

cos (x) = - \frac{3 + 2}{2}

cos (x) = - \frac{3 + 2}{2} :

حلول عامّة لـ

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{3 + 2}{2}) + 2 πn, x = - arccos (- \frac{3 + 2}{2}) + 2 πn

x = arccos (- \frac{3 + 2}{2}) + 2 πn, x = - arccos (- \frac{3 + 2}{2}) + 2 πn

وحّد الحلول

x = arccos (\frac{3 + 2}{2}) + 2 πn, x = 2 π - arccos (\frac{3 + 2}{2}) + 2 πn, x = arccos (- \frac{3 + 2}{2}) + 2 πn, x = - arccos (- \frac{3 + 2}{2}) + 2 πn

أظهر الحلّ بالتمثيل العشريّ

x = 0.26179 \dots + 2 πn, x = 2 π - 0.26179 \dots + 2 πn, x = 2.87979 \dots + 2 πn, x = - 2.87979 \dots + 2 πn

رسم

أمثلة شائعة

2tan^2(x)+cot^2(x)-3=0

((2sin(x)-1))/((sin(x)+5))=0

sec^2(b)=2+tan(b)

cos^{23}(x)+cos^2(x)=0

sin(b)=0.775