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شائع علم المثلثات >

8sin^4(x)-6sin^2(x)+1=0

الحلّ

8 sin^{4} (x) - 6 sin^{2} (x) + 1 = 0

الحلّ

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 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 = 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\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

8 sin^{4} (x) - 6 sin^{2} (x) + 1 = 0

بالاستعانة بطريقة التعويض

8 sin^{4} (x) - 6 sin^{2} (x) + 1 = 0

sin (x) = u :

على افتراض أنّ

8 u^{4} - 6 u^{2} + 1 = 0

8 u^{4} - 6 u^{2} + 1 = 0 : u = \frac{1}{2}, u = - \frac{1}{2}, u = \frac{1}{2}, u = - \frac{1}{2}

8 u^{4} - 6 u^{2} + 1 = 0

v^{2} = u^{4}

وكذلك

v = u^{2}

اكتب المعادلة مجددًا، بحيث أنّ

8 v^{2} - 6 v + 1 = 0

8 v^{2} - 6 v + 1 = 0

حلّ:

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

8 v^{2} - 6 v + 1 = 0

حلّ بالاستعانة بالصيغة التربيعيّة

8 v^{2} - 6 v + 1 = 0

الصيغة لحلّ المعادلة التربيعيّة

a = 8, b = - 6, c = 1

لـ

v_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

v_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

(- 6)^{2} - 4 \cdot 8 \cdot 1 = 2

(- 6)^{2} - 4 \cdot 8 \cdot 1

زوجيّ n

إذا تحقّق أنّ

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

:فعّل قانون القوى

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 8 \cdot 1

4 \cdot 8 \cdot 1 = 32 :

اضرب الأعداد

= 6^{2} - 32

6^{2} = 36

= 36 - 32

36 - 32 = 4 :

اطرح الأعداد

= 4

4 = 2^{2} :

حلّل العدد لعوامله أوّليّة

= 2^{2}

n a^{n} = a

:فعْل قانون الجذور

2^{2} = 2

= 2

v_{1, 2} = \frac{- ( - 6 ) \pm 2}{2 \cdot 8}

Separate the solutions

v_{1} = \frac{- ( - 6 ) + 2}{2 \cdot 8}, v_{2} = \frac{- ( - 6 ) - 2}{2 \cdot 8}

v = \frac{- ( - 6 ) + 2}{2 \cdot 8} : \frac{1}{2}

\frac{- ( - 6 ) + 2}{2 \cdot 8}

- (- a) = a

فعّل القانون

= \frac{6 + 2}{2 \cdot 8}

6 + 2 = 8 :

اجمع الأعداد

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

2 \cdot 8 = 16 :

اضرب الأعداد

= \frac{8}{16}

8 :

إلغ العوامل المشتركة

= \frac{1}{2}

v = \frac{- ( - 6 ) - 2}{2 \cdot 8} : \frac{1}{4}

\frac{- ( - 6 ) - 2}{2 \cdot 8}

- (- a) = a

فعّل القانون

= \frac{6 - 2}{2 \cdot 8}

6 - 2 = 4 :

اطرح الأعداد

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

2 \cdot 8 = 16 :

اضرب الأعداد

= \frac{4}{16}

4 :

إلغ العوامل المشتركة

= \frac{1}{4}

حلول المعادلة التربيعيّة هي

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

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

Substitute back

v = u^{2},

solve for

u

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

حلّ:

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

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

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

الحلول هي

x^{2} = f (a)

لـ

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 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}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 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{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 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}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 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^2(x)+2cos(x)+a=3

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

cos^5(x)+cos(x)+4cos^2(x)=2

cos^{5} (x) + cos (x) + 4 cos^{2} (x) = 2

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

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

(sin^3(x))/(2+2(sin(x))^2)=1

\frac{sin ^{3} ( x )}{2 + 2 ( sin ( x ) ) ^{2}} = 1

solvefor x,cot^2(x)= 1/3

so l v e f or x, cot^{2} (x) = \frac{1}{3}