English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

شائع علم المثلثات >

solvefor x,sin(x)=a-2qcos(2x)

الحلّ

solve for

x, sin (x) = a - 2 \cdot q \cdot cos (2 x)

الحلّ

x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

خطوات الحلّ

sin (x) = a - 2 q cos (2 x)

من الطرفين

a - 2 q cos (2 x)

اطرح

sin (x) - a + 2 q cos (2 x) = 0

Rewrite using trig identities

sin (x) - a + 2 cos (2 x) q

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

:فعّل متطابقة الزاوية المضاعفة

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

sin (x) - a + (1 - 2 sin^{2} (x)) \cdot 2 q = 0

بالاستعانة بطريقة التعويض

sin (x) - a + (1 - 2 sin^{2} (x)) \cdot 2 q = 0

sin (x) = u :

على افتراض أنّ

u - a + (1 - 2 u^{2}) \cdot 2 q = 0

u - a + (1 - 2 u^{2}) \cdot 2 q = 0 : u = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}, u = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}; q \neq = 0

u - a + (1 - 2 u^{2}) \cdot 2 q = 0

u - a + (1 - 2 u^{2}) \cdot 2 q

وسّع:

u - a + 2 q - 4 q u^{2}

u - a + (1 - 2 u^{2}) \cdot 2 q

= u - a + 2 q (1 - 2 u^{2})

2 q (1 - 2 u^{2})

وسٌع:

2 q - 4 q u^{2}

2 q (1 - 2 u^{2})

a (b - c) = ab - a c

: افتح أقواس بالاستعانة بـ

a = 2 q, b = 1, c = 2 u^{2}

= 2 q \cdot 1 - 2 q \cdot 2 u^{2}

= 2 \cdot 1 \cdot q - 2 \cdot 2 q u^{2}

2 \cdot 1 \cdot q - 2 \cdot 2 q u^{2}

بسّط:

2 q - 4 q u^{2}

2 \cdot 1 \cdot q - 2 \cdot 2 q u^{2}

2 \cdot 1 = 2 :

اضرب الأعداد

= 2 q - 2 \cdot 2 q u^{2}

2 \cdot 2 = 4 :

اضرب الأعداد

= 2 q - 4 q u^{2}

= 2 q - 4 q u^{2}

= u - a + 2 q - 4 q u^{2}

u - a + 2 q - 4 q u^{2} = 0

a x^{2} + b x + c = 0

اكتب بالصورة الاعتياديّة

- 4 q u^{2} + u - a + 2 q = 0

حلّ بالاستعانة بالصيغة التربيعيّة

- 4 q u^{2} + u - a + 2 q = 0

الصيغة لحلّ المعادلة التربيعيّة

a = - 4 q, b = 1, c = - a + 2 q

لـ

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 4 q ) ( - a + 2 q )}{2 ( - 4 q )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 4 q ) ( - a + 2 q )}{2 ( - 4 q )}

1^{2} - 4 (- 4 q) (- a + 2 q)

بسّط:

1 + 16 q (- a + 2 q)

1^{2} - 4 (- 4 q) (- a + 2 q)

1^{a} = 1

فعّل القانون

1^{2} = 1

= 1 - 4 (- 4 q) (2 q - a)

- (- a) = a

فعّل القانون

= 1 + 4 \cdot 4 q (- a + 2 q)

4 \cdot 4 = 16 :

اضرب الأعداد

= 1 + 16 q (2 q - a)

u_{1, 2} = \frac{- 1 \pm 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )}; q \neq = 0

Separate the solutions

u_{1} = \frac{- 1 + 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )}, u_{2} = \frac{- 1 - 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )}

u = \frac{- 1 + 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )} : - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}

\frac{- 1 + 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )}

(- a) = - a

:احذف الأقواس

= \frac{- 1 + 1 + 16 q ( - a + 2 q )}{- 2 \cdot 4 q}

2 \cdot 4 = 8 :

اضرب الأعداد

= \frac{- 1 + 16 q ( 2 q - a ) + 1}{- 8 q}

\frac{a}{- b} = - \frac{a}{b}

: استخدم ميزات الكسور التالية

= - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}

u = \frac{- 1 - 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )} : - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}

\frac{- 1 - 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )}

(- a) = - a

:احذف الأقواس

= \frac{- 1 - 1 + 16 q ( - a + 2 q )}{- 2 \cdot 4 q}

2 \cdot 4 = 8 :

اضرب الأعداد

= \frac{- 1 - 16 q ( 2 q - a ) + 1}{- 8 q}

\frac{a}{- b} = - \frac{a}{b}

: استخدم ميزات الكسور التالية

= - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}

حلول المعادلة التربيعيّة هي

u = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}, u = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}; q \neq = 0

u = sin (x)

استبدل مجددًا

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}, sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}; q \neq = 0

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}, sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}; q \neq = 0

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q} : x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}

Apply trig inverse properties

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q} :

حلول عامّة لـ

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q} : x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}

Apply trig inverse properties

sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}

sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q} :

حلول عامّة لـ

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

وحّد الحلول

x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

رسم

أمثلة شائعة

cos(2x)-3sin(-x)+2=0

cos (2 x) - 3 sin (- x) + 2 = 0

0=-sin(8x)+sin(6x)

0 = - sin (8 x) + sin (6 x)

3tan(x)sin(x)-sqrt(3)sin(x)=0

3 tan (x) sin (x) - 3 sin (x) = 0

cot^4(x)-3cot^2(x)+1=0

cot^{4} (x) - 3 cot^{2} (x) + 1 = 0

(cos(-30+x))/(cos(30+x))= 1835/726

\frac{cos ( - 3 0 ^{\circ} + x )}{cos ( 3 0 ^{\circ} + x )} = \frac{1835}{726}

solvefor x,sin(x)=a-2*q*cos(2x)

الحلّ

الحلّ

رسم

أمثلة شائعة

solvefor x,sin(x)=a-2qcos(2x)