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شائع علم المثلثات >

-arccos(x)+arctan(5/12)=0

الحلّ

- arccos (x) + arctan (\frac{5}{12}) = 0

الحلّ

x = \frac{12}{13}

خطوات الحلّ

- arccos (x) + arctan (\frac{5}{12}) = 0

انقل

arctan (\frac{5}{12})

إلى الجانب الأيمن

- arccos (x) + arctan (\frac{5}{12}) = 0

من الطرفين

arctan (\frac{5}{12})

اطرح

- arccos (x) + arctan (\frac{5}{12}) - arctan (\frac{5}{12}) = 0 - arctan (\frac{5}{12})

بسّط

- arccos (x) = - arctan (\frac{5}{12})

- arccos (x) = - arctan (\frac{5}{12})

- 1

اقسم الطرفين على

- arccos (x) = - arctan (\frac{5}{12})

- 1

اقسم الطرفين على

\frac{- arccos ( x )}{- 1} = \frac{- arctan ( \frac{5}{12} )}{- 1}

بسّط

\frac{- arccos ( x )}{- 1} = \frac{- arctan ( \frac{5}{12} )}{- 1}

\frac{- arccos ( x )}{- 1}

بسّط:

arccos (x)

\frac{- arccos ( x )}{- 1}

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

: استخدم ميزات الكسور التالية

= \frac{arccos ( x )}{1}

\frac{a}{1} = a

فعّل القانون

= arccos (x)

\frac{- arctan ( \frac{5}{12} )}{- 1}

بسّط:

arctan (\frac{5}{12})

\frac{- arctan ( \frac{5}{12} )}{- 1}

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

: استخدم ميزات الكسور التالية

= \frac{arctan ( \frac{5}{12} )}{1}

\frac{a}{1} = a

: استخدم ميزات الكسور التالية

= arctan (\frac{5}{12})

arccos (x) = arctan (\frac{5}{12})

arccos (x) = arctan (\frac{5}{12})

arccos (x) = arctan (\frac{5}{12})

Apply trig inverse properties

arccos (x) = arctan (\frac{5}{12})

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

x = cos (arctan (\frac{5}{12}))

cos (arctan (\frac{5}{12})) = \frac{12}{13}

cos (arctan (\frac{5}{12}))

Rewrite using trig identities:

cos (arctan (\frac{5}{12})) = \frac{1 + ( \frac{5}{12} ) ^{2}}{1 + ( \frac{5}{12} ) ^{2}}

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

استخدم المتطابقة التالية

= \frac{1 + ( \frac{5}{12} ) ^{2}}{1 + ( \frac{5}{12} ) ^{2}}

= \frac{1 + ( \frac{5}{12} ) ^{2}}{1 + ( \frac{5}{12} ) ^{2}}

بسّط

= \frac{12}{13}

x = \frac{12}{13}

x = \frac{12}{13}

رسم

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