What is the general solution for tan(x)-tan(pi/(18))=1+tan(x)tan(pi/(18)) ?

The general solution for tan(x)-tan(pi/(18))=1+tan(x)tan(pi/(18)) is x=0.95993…+pin

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Popular >

tan(x)-tan(pi/(18))=1+tan(x)tan(pi/(18))

Solution

tan (x) - tan (\frac{π}{18}) = 1 + tan (x) tan (\frac{π}{18})

Solution

x = 0.95993 \dots + πn

Degrees

x = 5 5^{\circ} + 18 0^{\circ} n

Solution steps

tan (x) - tan (\frac{π}{18}) = 1 + tan (x) tan (\frac{π}{18})

Solve by substitution

tan (x) - tan (\frac{π}{18}) = 1 + tan (x) tan (\frac{π}{18})

Let:

tan (x) = u

u - tan (\frac{π}{18}) = 1 + u tan (\frac{π}{18})

u - tan (\frac{π}{18}) = 1 + u tan (\frac{π}{18}) : u = \frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}

u - tan (\frac{π}{18}) = 1 + u tan (\frac{π}{18})

Move

tan (\frac{π}{18})

to the right side

u - tan (\frac{π}{18}) = 1 + u tan (\frac{π}{18})

Add

tan (\frac{π}{18})

to both sides

u - tan (\frac{π}{18}) + tan (\frac{π}{18}) = 1 + u tan (\frac{π}{18}) + tan (\frac{π}{18})

Simplify

u = 1 + u tan (\frac{π}{18}) + tan (\frac{π}{18})

u = 1 + u tan (\frac{π}{18}) + tan (\frac{π}{18})

Move

u tan (\frac{π}{18})

to the left side

u = 1 + u tan (\frac{π}{18}) + tan (\frac{π}{18})

Subtract

u tan (\frac{π}{18})

from both sides

u - u tan (\frac{π}{18}) = 1 + u tan (\frac{π}{18}) + tan (\frac{π}{18}) - u tan (\frac{π}{18})

Simplify

u - u tan (\frac{π}{18}) = 1 + tan (\frac{π}{18})

u - u tan (\frac{π}{18}) = 1 + tan (\frac{π}{18})

Factor

u - u tan (\frac{π}{18}) : (1 - tan (\frac{π}{18})) u

u - u tan (\frac{π}{18})

Factor out common term

u

= u (1 - tan (\frac{π}{18}))

(1 - tan (\frac{π}{18})) u = 1 + tan (\frac{π}{18})

Divide both sides by

1 - tan (\frac{π}{18})

(1 - tan (\frac{π}{18})) u = 1 + tan (\frac{π}{18})

Divide both sides by

1 - tan (\frac{π}{18})

\frac{( 1 - tan ( \frac{π}{18} ) ) u}{1 - tan ( \frac{π}{18} )} = \frac{1}{1 - tan ( \frac{π}{18} )} + \frac{tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}

Simplify

\frac{( 1 - tan ( \frac{π}{18} ) ) u}{1 - tan ( \frac{π}{18} )} = \frac{1}{1 - tan ( \frac{π}{18} )} + \frac{tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}

Simplify

\frac{( 1 - tan ( \frac{π}{18} ) ) u}{1 - tan ( \frac{π}{18} )} : u

\frac{( 1 - tan ( \frac{π}{18} ) ) u}{1 - tan ( \frac{π}{18} )}

Cancel the common factor:

1 - tan (\frac{π}{18})

= u

Simplify

\frac{1}{1 - tan ( \frac{π}{18} )} + \frac{tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )} : \frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}

\frac{1}{1 - tan ( \frac{π}{18} )} + \frac{tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}

Apply rule

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

= \frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}

u = \frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}

u = \frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}

u = \frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}

Substitute back

u = tan (x)

tan (x) = \frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}

tan (x) = \frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}

tan (x) = \frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )} : x = arctan (\frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}) + πn

tan (x) = \frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}

Apply trig inverse properties

tan (x) = \frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}

General solutions for

tan (x) = \frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}) + πn

x = arctan (\frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}) + πn

Combine all the solutions

x = arctan (\frac{1 + tan ( \frac{π}{18} )}{1 - tan ( \frac{π}{18} )}) + πn

Show solutions in decimal form

x = 0.95993 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan(x)-tan(pi/(18))=1+tan(x)tan(pi/(18)) ?
The general solution for tan(x)-tan(pi/(18))=1+tan(x)tan(pi/(18)) is x=0.95993…+pin