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شائع >

-sin^3(t)+sin(2t)cos(t)=0

الحلّ

- sin^{3} (t) + sin (2 t) cos (t) = 0

الحلّ

t = 2 πn, t = π + 2 πn, t = - 0.95531 \dots + πn, t = 0.95531 \dots + πn

درجات

t = 0^{\circ} + 36 0^{\circ} n, t = 18 0^{\circ} + 36 0^{\circ} n, t = - 54.73561 \dots^{\circ} + 18 0^{\circ} n, t = 54.73561 \dots^{\circ} + 18 0^{\circ} n

خطوات الحلّ

- sin^{3} (t) + sin (2 t) cos (t) = 0

Rewrite using trig identities

- sin^{3} (t) + cos (t) sin (2 t)

sin (2 x) = 2 sin (x) cos (x)

:فعّل متطابقة الزاوية المضاعفة

= - sin^{3} (t) + cos (t) \cdot 2 sin (t) cos (t)

cos (t) \cdot 2 sin (t) cos (t) = 2 cos^{2} (t) sin (t)

cos (t) \cdot 2 sin (t) cos (t)

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

:فعّل قانون القوى

cos (t) cos (t) = cos^{1 + 1} (t)

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

1 + 1 = 2 :

اجمع الأعداد

= 2 sin (t) cos^{2} (t)

= - sin^{3} (t) + 2 cos^{2} (t) sin (t)

- sin^{3} (t) + 2 cos^{2} (t) sin (t) = 0

- sin^{3} (t) + 2 cos^{2} (t) sin (t)

حلل إلى عوامل:

sin (t) (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

- sin^{3} (t) + 2 cos^{2} (t) sin (t)

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

:فعّل قانون القوى

sin^{3} (t) = sin (t) sin^{2} (t)

= - sin (t) sin^{2} (t) + 2 sin (t) cos^{2} (t)

sin (t)

قم باخراج العامل المشترك

= sin (t) (- sin^{2} (t) + 2 cos^{2} (t))

2 cos^{2} (t) - sin^{2} (t)

حلل إلى عوامل:

(2 cos (t) + sin (t)) (2 cos (t) - sin (t))

2 cos^{2} (t) - sin^{2} (t)

(2 cos (t))^{2} - sin^{2} (t)

كـ

2 cos^{2} (t) - sin^{2} (t)

اكتب مجددًا

2 cos^{2} (t) - sin^{2} (t)

a = (a)^{2}

:فعْل قانون الجذور

2 = (2)^{2}

= (2)^{2} cos^{2} (t) - sin^{2} (t)

a^{m} b^{m} = (ab)^{m}

:فعّل قانون القوى

(2)^{2} cos^{2} (t) = (2 cos (t))^{2}

= (2 cos (t))^{2} - sin^{2} (t)

= (2 cos (t))^{2} - sin^{2} (t)

x^{2} - y^{2} = (x + y) (x - y)

فعّل قانون فرق المربّعات

(2 cos (t))^{2} - sin^{2} (t) = (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

= (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

= sin (t) (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

sin (t) (2 cos (t) + sin (t)) (2 cos (t) - sin (t)) = 0

حلّ كل جزء على حدة

sin (t) = 0 or 2 cos (t) + sin (t) = 0 or 2 cos (t) - sin (t) = 0

sin (t) = 0 : t = 2 πn, t = π + 2 πn

sin (t) = 0

sin (t) = 0 :

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}

t = 0 + 2 πn, t = π + 2 πn

t = 0 + 2 πn, t = π + 2 πn

t = 0 + 2 πn

حلّ:

t = 2 πn

t = 0 + 2 πn

0 + 2 πn = 2 πn

t = 2 πn

t = 2 πn, t = π + 2 πn

2 cos (t) + sin (t) = 0 : t = arctan (- 2) + πn

2 cos (t) + sin (t) = 0

Rewrite using trig identities

2 cos (t) + sin (t) = 0

cos (t) \neq = 0, cos (t)

اقسم الطرفين على

\frac{2 cos ( t ) + sin ( t )}{cos ( t )} = \frac{0}{cos ( t )}

بسّط

2 + \frac{sin ( t )}{cos ( t )} = 0

\frac{sin ( x )}{cos ( x )} = tan (x)

:Use the basic trigonometric identity

2 + tan (t) = 0

2 + tan (t) = 0

انقل

2

إلى الجانب الأيمن

2 + tan (t) = 0

من الطرفين

2

اطرح

2 + tan (t) - 2 = 0 - 2

بسّط

tan (t) = - 2

tan (t) = - 2

Apply trig inverse properties

tan (t) = - 2

tan (t) = - 2 :

حلول عامّة لـ

tan (x) = - a \Rightarrow x = arctan (- a) + πn

t = arctan (- 2) + πn

t = arctan (- 2) + πn

2 cos (t) - sin (t) = 0 : t = arctan (2) + πn

2 cos (t) - sin (t) = 0

Rewrite using trig identities

2 cos (t) - sin (t) = 0

cos (t) \neq = 0, cos (t)

اقسم الطرفين على

\frac{2 cos ( t ) - sin ( t )}{cos ( t )} = \frac{0}{cos ( t )}

بسّط

2 - \frac{sin ( t )}{cos ( t )} = 0

\frac{sin ( x )}{cos ( x )} = tan (x)

:Use the basic trigonometric identity

2 - tan (t) = 0

2 - tan (t) = 0

انقل

2

إلى الجانب الأيمن

2 - tan (t) = 0

من الطرفين

2

اطرح

2 - tan (t) - 2 = 0 - 2

بسّط

- tan (t) = - 2

- tan (t) = - 2

- 1

اقسم الطرفين على

- tan (t) = - 2

- 1

اقسم الطرفين على

\frac{- tan ( t )}{- 1} = \frac{- 2}{- 1}

بسّط

tan (t) = 2

tan (t) = 2

Apply trig inverse properties

tan (t) = 2

tan (t) = 2 :

حلول عامّة لـ

tan (x) = a \Rightarrow x = arctan (a) + πn

t = arctan (2) + πn

t = arctan (2) + πn

وحّد الحلول

t = 2 πn, t = π + 2 πn, t = arctan (- 2) + πn, t = arctan (2) + πn

أظهر الحلّ بالتمثيل العشريّ

t = 2 πn, t = π + 2 πn, t = - 0.95531 \dots + πn, t = 0.95531 \dots + πn

رسم

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