What is the general solution for (tan(x)-1)/(1+tan(x))=0 ?

The general solution for (tan(x)-1)/(1+tan(x))=0 is x= pi/4+pin

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Popular >

(tan(x)-1)/(1+tan(x))=0

Solution

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

Solution

x = \frac{π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

tan (x) = u

\frac{u - 1}{1 + u} = 0

\frac{u - 1}{1 + u} = 0 : u = 1

\frac{u - 1}{1 + u} = 0

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

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

Verify Solutions

Find undefined (singularity) points:

u = - 1

Take the denominator(s) of

\frac{u - 1}{1 + u}

and compare to zero

Solve

1 + u = 0 : u = - 1

1 + u = 0

Move

1

to the right side

1 + u = 0

Subtract

1

from both sides

1 + u - 1 = 0 - 1

Simplify

u = - 1

u = - 1

The following points are undefined

u = - 1

Combine undefined points with solutions:

u = 1

Substitute back

u = tan (x)

tan (x) = 1

tan (x) = 1

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{π}{4} + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for (tan(x)-1)/(1+tan(x))=0 ?
The general solution for (tan(x)-1)/(1+tan(x))=0 is x= pi/4+pin