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(tan(x)+cos(x))/(1+sin(x))=sec(x)

الحلّ

\frac{tan ( x ) + cos ( x )}{1 + sin ( x )} = sec (x)

الحلّ

x = 2 πn, x = π + 2 πn

درجات

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

خطوات الحلّ

\frac{tan ( x ) + cos ( x )}{1 + sin ( x )} = sec (x)

من الطرفين

sec (x)

اطرح

\frac{tan ( x ) + cos ( x )}{1 + sin ( x )} - sec (x) = 0

\frac{tan ( x ) + cos ( x )}{1 + sin ( x )} - sec (x)

بسّط:

\frac{tan ( x ) + cos ( x ) - sec ( x ) ( 1 + sin ( x ) )}{1 + sin ( x )}

\frac{tan ( x ) + cos ( x )}{1 + sin ( x )} - sec (x)

sec (x) = \frac{sec ( x ) ( 1 + sin ( x ) )}{1 + sin ( x )}

:حوّل الأعداد لكسور

= \frac{tan ( x ) + cos ( x )}{1 + sin ( x )} - \frac{sec ( x ) ( 1 + sin ( x ) )}{1 + sin ( x )}

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

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{tan ( x ) + cos ( x ) - sec ( x ) ( 1 + sin ( x ) )}{1 + sin ( x )}

\frac{tan ( x ) + cos ( x ) - sec ( x ) ( 1 + sin ( x ) )}{1 + sin ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

tan (x) + cos (x) - sec (x) (1 + sin (x)) = 0

sin, cos :

عبّر بواسطة

cos (x) + tan (x) - (1 + sin (x)) sec (x)

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

:Use the basic trigonometric identity

= cos (x) + \frac{sin ( x )}{cos ( x )} - (1 + sin (x)) sec (x)

sec (x) = \frac{1}{cos ( x )}

:Use the basic trigonometric identity

= cos (x) + \frac{sin ( x )}{cos ( x )} - (1 + sin (x)) \frac{1}{cos ( x )}

cos (x) + \frac{sin ( x )}{cos ( x )} - (1 + sin (x)) \frac{1}{cos ( x )}

بسّط:

\frac{cos ^{2} ( x ) - 1}{cos ( x )}

cos (x) + \frac{sin ( x )}{cos ( x )} - (1 + sin (x)) \frac{1}{cos ( x )}

(1 + sin (x)) \frac{1}{cos ( x )} = \frac{1 + sin ( x )}{cos ( x )}

(1 + sin (x)) \frac{1}{cos ( x )}

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

:اضرب كسور

= \frac{1 \cdot ( 1 + sin ( x ) )}{cos ( x )}

1 \cdot (1 + sin (x)) = 1 + sin (x)

1 \cdot (1 + sin (x))

1 \cdot (1 + sin (x)) = (1 + sin (x)) :

اضرب

= (1 + sin (x))

(a) = a

:احذف الأقواس

= 1 + sin (x)

= \frac{1 + sin ( x )}{cos ( x )}

= cos (x) + \frac{sin ( x )}{cos ( x )} - \frac{sin ( x ) + 1}{cos ( x )}

\frac{sin ( x )}{cos ( x )} - \frac{sin ( x ) + 1}{cos ( x )}

وحّد الكسور:

\frac{sin ( x ) - ( 1 + sin ( x ) )}{cos ( x )}

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

فعّل القانون

= \frac{sin ( x ) - ( sin ( x ) + 1 )}{cos ( x )}

= cos (x) + \frac{sin ( x ) - ( sin ( x ) + 1 )}{cos ( x )}

\frac{sin ( x ) - ( 1 + sin ( x ) )}{cos ( x )} = - \frac{1}{cos ( x )}

\frac{sin ( x ) - ( 1 + sin ( x ) )}{cos ( x )}

sin (x) - (1 + sin (x))

وسٌع:

- 1

sin (x) - (1 + sin (x))

- (1 + sin (x)) : - 1 - sin (x)

- (1 + sin (x))

افتح أقواس

= - (1) - (sin (x))

فعّل قوانين سالب-موجب

+ (- a) = - a

= - 1 - sin (x)

= sin (x) - 1 - sin (x)

sin (x) - 1 - sin (x)

بسّط:

- 1

sin (x) - 1 - sin (x)

جمّع التعابير المتشابهة

= sin (x) - sin (x) - 1

sin (x) - sin (x) = 0 :

اجمع العناصر المتشابهة

= - 1

= - 1

= \frac{- 1}{cos ( x )}

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

: استخدم ميزات الكسور التالية

= - \frac{1}{cos ( x )}

= cos (x) - \frac{1}{cos ( x )}

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

:حوّل الأعداد لكسور

= \frac{cos ( x ) cos ( x )}{cos ( x )} - \frac{1}{cos ( x )}

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

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{cos ( x ) cos ( x ) - 1}{cos ( x )}

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

cos (x) cos (x) - 1

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

cos (x) cos (x)

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

:فعّل قانون القوى

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

= cos^{1 + 1} (x)

1 + 1 = 2 :

اجمع الأعداد

= cos^{2} (x)

= cos^{2} (x) - 1

= \frac{cos ^{2} ( x ) - 1}{cos ( x )}

= \frac{cos ^{2} ( x ) - 1}{cos ( x )}

\frac{- 1 + cos ^{2} ( x )}{cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- 1 + cos^{2} (x) = 0

بالاستعانة بطريقة التعويض

- 1 + cos^{2} (x) = 0

cos (x) = u :

على افتراض أنّ

- 1 + u^{2} = 0

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

- 1 + u^{2} = 0

انقل

1

إلى الجانب الأيمن

- 1 + u^{2} = 0

للطرفين

1

أضف

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

بسّط

u^{2} = 1

u^{2} = 1

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

الحلول هي

x^{2} = f (a)

لـ

u = 1, u = - 1

1 = 1

1

1 = 1

فعّل القانون

= 1

- 1 = - 1

- 1

1 = 1

فعّل القانون

= - 1

u = 1, u = - 1

u = cos (x)

استبدل مجددًا

cos (x) = 1, cos (x) = - 1

cos (x) = 1, cos (x) = - 1

cos (x) = 1 : x = 2 πn

cos (x) = 1

cos (x) = 1 :

حلول عامّة لـ

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 2 πn

x = 0 + 2 πn

حلّ:

x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

cos (x) = - 1 :

حلول عامّة لـ

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

وحّد الحلول

x = 2 πn, x = π + 2 πn

أمثلة شائعة

sin^2(x)=2+2cos(x),0<= x<= 2pi

sin^{2} (x) = 2 + 2 cos (x), 0 \leq x \leq 2 π

cos(5x+3)= 1/2

cos (5 x + 3) = \frac{1}{2}

csc(θ)=3

csc (θ) = 3

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

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

-3-2cos(θ)=2

- 3 - 2 cos (θ) = 2