What is the general solution for arctan(4x-4)=-1 ?

The general solution for arctan(4x-4)=-1 is x=-(tan(1))/4+1

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arctan(4x-4)=-1

arctan (4 x - 4) = - 1

x = - \frac{tan ( 1 )}{4} + 1

Solution steps

arctan (4 x - 4) = - 1

Apply trig inverse properties

arctan (4 x - 4) = - 1

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

4 x - 4 = tan (- 1)

tan (- 1) = - tan (1)

tan (- 1)

Use the following property:

tan (- x) = - tan (x)

tan (- 1) = - tan (1)

= - tan (1)

4 x - 4 = - tan (1)

4 x - 4 = - tan (1)

Solve

4 x - 4 = - tan (1) : x = - \frac{tan ( 1 )}{4} + 1

4 x - 4 = - tan (1)

Move

4

to the right side

4 x - 4 = - tan (1)

Add

4

to both sides

4 x - 4 + 4 = - tan (1) + 4

Simplify

4 x = - tan (1) + 4

4 x = - tan (1) + 4

Divide both sides by

4

4 x = - tan (1) + 4

Divide both sides by

4

\frac{4 x}{4} = - \frac{tan ( 1 )}{4} + \frac{4}{4}

Simplify

\frac{4 x}{4} = - \frac{tan ( 1 )}{4} + \frac{4}{4}

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

- \frac{tan ( 1 )}{4} + \frac{4}{4} : - \frac{tan ( 1 )}{4} + 1

- \frac{tan ( 1 )}{4} + \frac{4}{4}

Apply rule

\frac{a}{a} = 1

\frac{4}{4} = 1

= - \frac{tan ( 1 )}{4} + 1

x = - \frac{tan ( 1 )}{4} + 1

x = - \frac{tan ( 1 )}{4} + 1

x = - \frac{tan ( 1 )}{4} + 1

x = - \frac{tan ( 1 )}{4} + 1

tan^{2} (x + \frac{π}{6}) = 3

cos (θ) cos (2 θ) + sin (θ) sin (2 θ) = \frac{3}{2}

3 cos (2 x) - 2 cos (x) - 1 = 0

sin (3 π x) = 0

tan (2 t) = 1, 0 \leq t < 2 π

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