What is the general solution for tan(θ)+1=0 ?

The general solution for tan(θ)+1=0 is θ=(3pi)/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan(θ)+1=0

tan (θ) + 1 = 0

θ = \frac{3 π}{4} + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

tan (θ) + 1 = 0

Move

1

to the right side

tan (θ) + 1 = 0

Subtract

1

from both sides

tan (θ) + 1 - 1 = 0 - 1

Simplify

tan (θ) = - 1

tan (θ) = - 1

General solutions for

tan (θ) = - 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}

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

arctan (\frac{π}{2})

sin (- \frac{3 π}{2})

arccos (1)

sec^{2} (x) - 4 = 0

sin (θ) = - \frac{2}{2}

What is the general solution for tan(θ)+1=0 ?
The general solution for tan(θ)+1=0 is θ=(3pi)/4+pin