What is the general solution for tan(2x-pi/4)=-1 ?

The general solution for tan(2x-pi/4)=-1 is x=(pin)/2+pi/2

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tan(2x-pi/4)=-1

tan (2 x - \frac{π}{4}) = - 1

x = \frac{πn}{2} + \frac{π}{2}

Degrees

x = 9 0^{\circ} + 9 0^{\circ} n

Solution steps

tan (2 x - \frac{π}{4}) = - 1

General solutions for

tan (2 x - \frac{π}{4}) = - 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}

2 x - \frac{π}{4} = \frac{3 π}{4} + πn

2 x - \frac{π}{4} = \frac{3 π}{4} + πn

Solve

2 x - \frac{π}{4} = \frac{3 π}{4} + πn : x = \frac{πn}{2} + \frac{π}{2}

2 x - \frac{π}{4} = \frac{3 π}{4} + πn

Move

\frac{π}{4}

to the right side

2 x - \frac{π}{4} = \frac{3 π}{4} + πn

Add

\frac{π}{4}

to both sides

2 x - \frac{π}{4} + \frac{π}{4} = \frac{3 π}{4} + πn + \frac{π}{4}

Simplify

2 x - \frac{π}{4} + \frac{π}{4} = \frac{3 π}{4} + πn + \frac{π}{4}

Simplify

2 x - \frac{π}{4} + \frac{π}{4} : 2 x

2 x - \frac{π}{4} + \frac{π}{4}

Add similar elements:

- \frac{π}{4} + \frac{π}{4} = 0

= 2 x

Simplify

\frac{3 π}{4} + πn + \frac{π}{4} : πn + π

\frac{3 π}{4} + πn + \frac{π}{4}

Group like terms

= πn + \frac{π}{4} + \frac{3 π}{4}

Combine the fractions

\frac{π}{4} + \frac{3 π}{4} : π

Apply rule

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

= \frac{π + 3 π}{4}

Add similar elements:

π + 3 π = 4 π

= \frac{4 π}{4}

Divide the numbers:

\frac{4}{4} = 1

= π

= πn + π

2 x = πn + π

2 x = πn + π

2 x = πn + π

Divide both sides by

2

2 x = πn + π

Divide both sides by

2

\frac{2 x}{2} = \frac{πn}{2} + \frac{π}{2}

Simplify

x = \frac{πn}{2} + \frac{π}{2}

x = \frac{πn}{2} + \frac{π}{2}

x = \frac{πn}{2} + \frac{π}{2}

2 sin^{2} (A) = 5 sin (A) - 2

cot (θ) = \frac{12}{5}

θ, cos (θ) = - \frac{1}{2}

- 3 sin (b) - 2 = sin (b) - 2

5 sin (2 x) = 2 cos (2 x)

What is the general solution for tan(2x-pi/4)=-1 ?
The general solution for tan(2x-pi/4)=-1 is x=(pin)/2+pi/2