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solvefor t,x=arccos(1/(sqrt(1+t^2)))

الحلّ

solve for

t, x = arccos (\frac{1}{1 + t ^{2}})

الحلّ

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

خطوات الحلّ

x = arccos (\frac{1}{1 + t ^{2}})

بدّل الأطراف

arccos (\frac{1}{1 + t ^{2}}) = x

Apply trig inverse properties

arccos (\frac{1}{1 + t ^{2}}) = x

arccos (x) = a \Rightarrow x = cos (a)

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

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

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

حلّ:

t = \frac{1 - cos ^{2} ( x )}{cos ( x )} {x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn}, t = - \frac{1 - cos ^{2} ( x )}{cos ( x )} {x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn}

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

1 + t^{2}

اضرب الطرفين بـ

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

بسّط

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

بدّل الأطراف

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

cos (x)

اقسم الطرفين على

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

cos (x)

اقسم الطرفين على

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

بسّط

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

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

ربّع الطرفين:

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

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

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

(1 + t^{2})^{2}

وسّع:

1 + t^{2}

(1 + t^{2})^{2}

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

:فعْل قانون الجذور

= ((1 + t^{2})^{\frac{1}{2}})^{2}

(a^{b})^{c} = a^{b c}

:فعّل قانون القوى

= (1 + t^{2})^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

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

:اضرب كسور

= \frac{1 \cdot 2}{2}

2 :

إلغ العوامل المشتركة

= 1

= 1 + t^{2}

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

وسّع:

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

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

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

:فعّل قانون القوى

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

1^{a} = 1

فعّل القانون

1^{2} = 1

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

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

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

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

حلّ:

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

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

انقل

1

إلى الجانب الأيمن

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

من الطرفين

1

اطرح

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

بسّط

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

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

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

الحلول هي

x^{2} = f (a)

لـ

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

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

بسّط:

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

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

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

وحّد:

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

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

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

:حوّل الأعداد لكسور

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

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

اضرب

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

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

a \geq 0, b \geq 0

بافتراض أنّ

n \frac{a}{b} = \frac{n a}{n b}

:فعّل قانون الجذور

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

a \geq 0

بافتراض أنّ

n a^{n} = a

:فعّل قانون الجذور

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

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

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

بسّط:

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

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

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

وحّد:

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

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

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

:حوّل الأعداد لكسور

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

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

اضرب

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

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

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

بسّط:

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

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

a \geq 0, b \geq 0

بافتراض أنّ

n \frac{a}{b} = \frac{n a}{n b}

:فعّل قانون الجذور

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

a \geq 0

بافتراض أنّ

n a^{n} = a

:فعّل قانون الجذور

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

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

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

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

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

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

افحص الإجبات:

t = \frac{1 - cos ^{2} ( x )}{cos ( x )} {x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn}, t = - \frac{1 - cos ^{2} ( x )}{cos ( x )} {x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn}

للتحقّق من دقّة الحلول

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

عوّض الحلول في

إلغي الحلول التي تعطي قضيّة كذب

x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn \Leftarrow \frac{1}{1 + ( \frac{1 - c o s ^{2} ( x )}{c o s ( x )} ) ^{2}} = cos (x) : t = \frac{1 - cos ^{2} ( x )}{cos ( x )}

عوّض

\frac{1}{1 + ( \frac{1 - c o s ^{2} ( x )}{c o s ( x )} ) ^{2}} = cos (x)

بالاستعانة بطريقة التعويض

\frac{1}{1 + ( \frac{1 - c o s ^{2} ( x )}{c o s ( x )} ) ^{2}} = cos (x)

cos (x) = u :

على افتراض أنّ

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

\frac{1}{1 + ( \frac{1 - u ^{2}}{u} ) ^{2}} = u : u = 1

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

1 + (\frac{1 - u ^{2}}{u})^{2}

اضرب الطرفين بـ

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

بسّط

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

ربّع الطرفين:

1 = 1

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

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

1^{2}

وسّع:

1

1^{2}

1^{a} = 1

فعّل القانون

= 1

u 1 + (\frac{1 - u ^{2}}{u})^{2}^{2}

وسّع:

1

u 1 + (\frac{1 - u ^{2}}{u})^{2}^{2}

(a \cdot b)^{n} = a^{n} b^{n}

:فعّل قانون القوى

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

1 + (\frac{1 - u ^{2}}{u})^{2}^{2} : 1 + (\frac{1 - u ^{2}}{u})^{2}

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

:فعْل قانون الجذور

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

(a^{b})^{c} = a^{b c}

:فعّل قانون القوى

= 1 + (\frac{1 - u ^{2}}{u})^{2}^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

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

:اضرب كسور

= \frac{1 \cdot 2}{2}

2 :

إلغ العوامل المشتركة

= 1

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

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

1 + (\frac{1 - u ^{2}}{u})^{2} u^{2}

وسّع:

1

1 + (\frac{1 - u ^{2}}{u})^{2} u^{2}

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

(\frac{1 - u ^{2}}{u})^{2}

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

:فعّل قانون القوى

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

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

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

:فعْل قانون الجذور

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

(a^{b})^{c} = a^{b c}

:فعّل قانون القوى

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

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

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

:اضرب كسور

= \frac{1 \cdot 2}{2}

2 :

إلغ العوامل المشتركة

= 1

= 1 - u^{2}

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

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

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

a (b + c) = ab + a c

: افتح أقواس بالاستعانة بـ

a = u^{2}, b = 1, c = \frac{1 - u ^{2}}{u ^{2}}

= u^{2} \cdot 1 + u^{2} \frac{1 - u ^{2}}{u ^{2}}

= 1 \cdot u^{2} + \frac{1 - u ^{2}}{u ^{2}} u^{2}

1 \cdot u^{2} + \frac{1 - u ^{2}}{u ^{2}} u^{2}

بسّط:

1

1 \cdot u^{2} + \frac{1 - u ^{2}}{u ^{2}} u^{2}

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

1 \cdot u^{2}

1 \cdot u^{2} = u^{2} :

اضرب

= u^{2}

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

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

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

:اضرب كسور

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

u^{2} :

إلغ العوامل المشتركة

= 1 - u^{2}

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

جمّع التعابير المتشابهة

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

u^{2} - u^{2} = 0 :

اجمع العناصر المتشابهة

= 1

= 1

= 1

1 = 1

1 = 1

يتساوى الطرفان

يتحقّق لكلّ u

افحص الإجبات:

u < - 1

خطأ

, u = - 1

خطأ

, - 1 < u < 1

خطأ

, u = 1

صحيح

, u > 1

خطأ

للتحقّق من دقّة الحلول

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

عوّض الحلول في

إلغي الحلول التي تعطي قضيّة كذب

قضيّة كذب

\Leftarrow \frac{1}{1 + ( \frac{1 - u ^{2}}{u} ) ^{2}} \neq = u

u < - 1

عوّض

الحل للمعادلة هو

u = 1

u = cos (x)

استبدل مجددًا

cos (x) = 1

cos (x) = 1

cos (x) = u = 1 : x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

cos (x) = u = 1

Apply trig inverse properties

cos (x) = u = 1

cos (x) = u = 1 :

حلول عامّة لـ

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

وحّد الحلول

x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn \Leftarrow \frac{1}{1 + ( - \frac{1 - c o s ^{2} ( x )}{c o s ( x )} ) ^{2}} = cos (x) : t = - \frac{1 - cos ^{2} ( x )}{cos ( x )}

عوّض

\frac{1}{1 + ( - \frac{1 - c o s ^{2} ( x )}{c o s ( x )} ) ^{2}} = cos (x)

بالاستعانة بطريقة التعويض

\frac{1}{1 + ( - \frac{1 - c o s ^{2} ( x )}{c o s ( x )} ) ^{2}} = cos (x)

cos (x) = u :

على افتراض أنّ

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

\frac{1}{1 + ( - \frac{1 - u ^{2}}{u} ) ^{2}} = u : u = 1

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

1 + (\frac{1 - u ^{2}}{u})^{2}

اضرب الطرفين بـ

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

بسّط

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

\frac{1 + ( \frac{1 - u ^{2}}{u} ) ^{2}}{1 + ( - \frac{1 - u ^{2}}{u} ) ^{2}}

وسّع:

1

\frac{1 + ( \frac{1 - u ^{2}}{u} ) ^{2}}{1 + ( - \frac{1 - u ^{2}}{u} ) ^{2}}

\frac{x}{y} = \frac{x}{y}

:وحّد القوى

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

\frac{1 + ( \frac{1 - u ^{2}}{u} ) ^{2}}{1 + ( - \frac{1 - u ^{2}}{u} ) ^{2}}

وسّع:

1

\frac{1 + ( \frac{1 - u ^{2}}{u} ) ^{2}}{1 + ( - \frac{1 - u ^{2}}{u} ) ^{2}}

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

\frac{1 + ( \frac{1 - u ^{2}}{u} ) ^{2}}{1 + ( - \frac{1 - u ^{2}}{u} ) ^{2}}

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

1 + (- \frac{1 - u ^{2}}{u})^{2}

زوجيّ n

إذا تحقّق أنّ

, (- a)^{n} = a^{n}

:فعّل قانون القوى

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

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

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

\frac{a}{a} = 1

فعّل القانون

= 1

= 1

1 = 1

فعّل القانون

= 1

= 1

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

ربّع الطرفين:

1 = 1

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

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

1^{2}

وسّع:

1

1^{2}

1^{a} = 1

فعّل القانون

= 1

u 1 + (\frac{1 - u ^{2}}{u})^{2}^{2}

وسّع:

1

u 1 + (\frac{1 - u ^{2}}{u})^{2}^{2}

(a \cdot b)^{n} = a^{n} b^{n}

:فعّل قانون القوى

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

1 + (\frac{1 - u ^{2}}{u})^{2}^{2} : 1 + (\frac{1 - u ^{2}}{u})^{2}

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

:فعْل قانون الجذور

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

(a^{b})^{c} = a^{b c}

:فعّل قانون القوى

= 1 + (\frac{1 - u ^{2}}{u})^{2}^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

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

:اضرب كسور

= \frac{1 \cdot 2}{2}

2 :

إلغ العوامل المشتركة

= 1

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

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

1 + (\frac{1 - u ^{2}}{u})^{2} u^{2}

وسّع:

1

1 + (\frac{1 - u ^{2}}{u})^{2} u^{2}

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

(\frac{1 - u ^{2}}{u})^{2}

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

:فعّل قانون القوى

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

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

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

:فعْل قانون الجذور

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

(a^{b})^{c} = a^{b c}

:فعّل قانون القوى

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

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

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

:اضرب كسور

= \frac{1 \cdot 2}{2}

2 :

إلغ العوامل المشتركة

= 1

= 1 - u^{2}

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

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

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

a (b + c) = ab + a c

: افتح أقواس بالاستعانة بـ

a = u^{2}, b = 1, c = \frac{1 - u ^{2}}{u ^{2}}

= u^{2} \cdot 1 + u^{2} \frac{1 - u ^{2}}{u ^{2}}

= 1 \cdot u^{2} + \frac{1 - u ^{2}}{u ^{2}} u^{2}

1 \cdot u^{2} + \frac{1 - u ^{2}}{u ^{2}} u^{2}

بسّط:

1

1 \cdot u^{2} + \frac{1 - u ^{2}}{u ^{2}} u^{2}

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

1 \cdot u^{2}

1 \cdot u^{2} = u^{2} :

اضرب

= u^{2}

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

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

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

:اضرب كسور

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

u^{2} :

إلغ العوامل المشتركة

= 1 - u^{2}

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

جمّع التعابير المتشابهة

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

u^{2} - u^{2} = 0 :

اجمع العناصر المتشابهة

= 1

= 1

= 1

1 = 1

1 = 1

يتساوى الطرفان

يتحقّق لكلّ u

افحص الإجبات:

u < - 1

خطأ

, u = - 1

خطأ

, - 1 < u < 1

خطأ

, u = 1

صحيح

, u > 1

خطأ

للتحقّق من دقّة الحلول

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

عوّض الحلول في

إلغي الحلول التي تعطي قضيّة كذب

قضيّة كذب

\Leftarrow \frac{1}{1 + ( - \frac{1 - u ^{2}}{u} ) ^{2}} \neq = u

u < - 1

عوّض

الحل للمعادلة هو

u = 1

u = cos (x)

استبدل مجددًا

cos (x) = 1

cos (x) = 1

cos (x) = u = 1 : x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

cos (x) = u = 1

Apply trig inverse properties

cos (x) = u = 1

cos (x) = u = 1 :

حلول عامّة لـ

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

وحّد الحلول

x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

The solutions are

t = \frac{1 - cos ^{2} ( x )}{cos ( x )} {x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn}, t = - \frac{1 - cos ^{2} ( x )}{cos ( x )} {x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn}

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

رسم

أمثلة شائعة

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

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

2cos^2(x)+15sin(x)-15=0

2 cos^{2} (x) + 15 sin (x) - 15 = 0

cos(2x)=sin(x),-2pi<= x<= 2pi

cos (2 x) = sin (x), - 2 π \leq x \leq 2 π

5sec^2(θ)sin(θ)-cos(θ)=0

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

cos(x)= 8/11

cos (x) = \frac{8}{11}