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

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

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cot(θ)+1=0,(0,2pi)

cot (θ) + 1 = 0, (0, 2 π)

θ = \frac{3 π}{4}, θ = \frac{7 π}{4}

Degrees

θ = 13 5^{\circ}, θ = 31 5^{\circ}

Solution steps

cot (θ) + 1 = 0, (0, 2 π)

Move

1

to the right side

cot (θ) + 1 = 0

Subtract

1

from both sides

cot (θ) + 1 - 1 = 0 - 1

Simplify

cot (θ) = - 1

cot (θ) = - 1

General solutions for

cot (θ) = - 1

cot (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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

Solutions for the range

(0, 2 π)

θ = \frac{3 π}{4}, θ = \frac{7 π}{4}

cos (θ) = \frac{4}{3.3 8 ^{2} + 4 ^{2}}

tan (x - 1 0^{\circ}) = - 0.1

2 (cos (x) + 3) = 5 + cos (x)

4 cos (3 x) - 1 = 2 sin (3 x) tan (3 x)

tan (θ) = \frac{15}{20}

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