What is the general solution for tan(θ)+1=0,0<= θ<= 2pi ?

The general solution for tan(θ)+1=0,0<= θ<= 2pi is θ=(3pi)/4 ,θ=(7pi)/4

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tan(θ)+1=0,0<= θ<= 2pi

tan (θ) + 1 = 0, 0 \leq θ \leq 2 π

θ = \frac{3 π}{4}, θ = \frac{7 π}{4}

Degrees

θ = 13 5^{\circ}, θ = 31 5^{\circ}

Solution steps

tan (θ) + 1 = 0, 0 \leq θ \leq 2 π

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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{3 π}{4}, θ = \frac{7 π}{4}

2 - sin (2 θ) = 4 sin (2 θ)

2 sin (θ) cos (θ) + sin (θ) = 0

csc (3 x) - 8 = 0

sin^{2} (x) = tan (x)

sin (x) \cdot cos (x) = \frac{1}{2}

What is the general solution for tan(θ)+1=0,0<= θ<= 2pi ?
The general solution for tan(θ)+1=0,0<= θ<= 2pi is θ=(3pi)/4 ,θ=(7pi)/4