What is the general solution for -arccos(x)+arctan(5/12)=0 ?

The general solution for -arccos(x)+arctan(5/12)=0 is x= 12/13

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

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

x = \frac{12}{13}

Solution steps

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

Move

arctan (\frac{5}{12})

to the right side

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

Subtract

arctan (\frac{5}{12})

from both sides

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

Simplify

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

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

Divide both sides by

- 1

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

Divide both sides by

- 1

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

Simplify

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

Simplify

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= arccos (x)

Simplify

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

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

Apply the fraction rule:

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

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

Apply the fraction rule:

\frac{a}{1} = a

= 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}}

Use the following identity:

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

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

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

Simplify

= \frac{12}{13}

x = \frac{12}{13}

x = \frac{12}{13}

tan (x) + cot (x) = 1

3 cos (x) + sin (x) = 1

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

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

sin (x) + ∣ sin (x) ∣ = 1

What is the general solution for -arccos(x)+arctan(5/12)=0 ?
The general solution for -arccos(x)+arctan(5/12)=0 is x= 12/13