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9(2sin^2(x)-1)=0

الحلّ

9 (2 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

+1

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

خطوات الحلّ

9 (2 sin^{2} (x) - 1) = 0

بالاستعانة بطريقة التعويض

9 (2 sin^{2} (x) - 1) = 0

sin (x) = u :

على افتراض أنّ

9 (2 u^{2} - 1) = 0

9 (2 u^{2} - 1) = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

9 (2 u^{2} - 1) = 0

9

اقسم الطرفين على

9 (2 u^{2} - 1) = 0

9

اقسم الطرفين على

\frac{9 ( 2 u ^{2} - 1 )}{9} = \frac{0}{9}

بسّط

2 u^{2} - 1 = 0

2 u^{2} - 1 = 0

انقل

1

إلى الجانب الأيمن

2 u^{2} - 1 = 0

للطرفين

1

أضف

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

بسّط

2 u^{2} = 1

2 u^{2} = 1

2

اقسم الطرفين على

2 u^{2} = 1

2

اقسم الطرفين على

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

بسّط

u^{2} = \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 = 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}{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

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

أمثلة شائعة

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

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

sin(9x)-cos(x)=0

sin (9 x) - cos (x) = 0

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

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

solvefor t,0.08=0.1cos(4t)solve for

t, 0.08 = 0.1 cos (4 t)

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