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شائع علم المثلثات >

tan(x)= 1/(sec(x))

الحلّ

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

الحلّ

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

درجات

x = 38.17270 \dots^{\circ} + 36 0^{\circ} n, x = 141.82729 \dots^{\circ} + 36 0^{\circ} n

خطوات الحلّ

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

من الطرفين

\frac{1}{sec ( x )}

اطرح

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

tan (x) - \frac{1}{sec ( x )}

بسّط:

\frac{tan ( x ) sec ( x ) - 1}{sec ( x )}

tan (x) - \frac{1}{sec ( x )}

tan (x) = \frac{tan ( x ) sec ( x )}{sec ( x )}

:حوّل الأعداد لكسور

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

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

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

tan (x) sec (x) - 1 = 0

sin, cos :

عبّر بواسطة

- 1 + sec (x) tan (x)

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

:Use the basic trigonometric identity

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

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

:Use the basic trigonometric identity

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

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

بسّط:

\frac{- cos ^{2} ( x ) + sin ( x )}{cos ^{2} ( x )}

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

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

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

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

:اضرب كسور

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

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

اضرب

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

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)

= \frac{sin ( x )}{cos ^{2} ( x )}

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

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

:حوّل الأعداد لكسور

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

1 \cdot cos^{2} (x) = cos^{2} (x) :

اضرب

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

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

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

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

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

Rewrite using trig identities

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

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

:فعّل نطريّة فيتاغوروس

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

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

- (1 - sin^{2} (x)) : - 1 + sin^{2} (x)

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

افتح أقواس

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

فعّل قوانين سالب-موجب

- (- a) = a, - (a) = - a

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

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

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

بالاستعانة بطريقة التعويض

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

sin (x) = u :

على افتراض أنّ

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

- 1 + u + u^{2} = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

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

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

اكتب بالصورة الاعتياديّة

u^{2} + u - 1 = 0

حلّ بالاستعانة بالصيغة التربيعيّة

u^{2} + u - 1 = 0

الصيغة لحلّ المعادلة التربيعيّة

a = 1, b = 1, c = - 1

لـ

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

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

1^{a} = 1

فعّل القانون

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

- (- a) = a

فعّل القانون

= 1 + 4 \cdot 1 \cdot 1

4 \cdot 1 \cdot 1 = 4 :

اضرب الأعداد

= 1 + 4

1 + 4 = 5 :

اجمع الأعداد

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 \cdot 1}, u_{2} = \frac{- 1 - 5}{2 \cdot 1}

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

2 \cdot 1 = 2 :

اضرب الأعداد

= \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

\frac{- 1 - 5}{2 \cdot 1}

2 \cdot 1 = 2 :

اضرب الأعداد

= \frac{- 1 - 5}{2}

حلول المعادلة التربيعيّة هي

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

u = sin (x)

استبدل مجددًا

sin (x) = \frac{- 1 + 5}{2}, sin (x) = \frac{- 1 - 5}{2}

sin (x) = \frac{- 1 + 5}{2}, sin (x) = \frac{- 1 - 5}{2}

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

sin (x) = \frac{- 1 + 5}{2}

Apply trig inverse properties

sin (x) = \frac{- 1 + 5}{2}

sin (x) = \frac{- 1 + 5}{2} :

حلول عامّة لـ

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

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

sin (x) = \frac{- 1 - 5}{2} :

لا يوجد حلّ

sin (x) = \frac{- 1 - 5}{2}

- 1 \leq sin (x) \leq 1

لا يوجد حلّ

وحّد الحلول

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

أظهر الحلّ بالتمثيل العشريّ

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

رسم

أمثلة شائعة

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