What is the general solution for (cot(θ)+1)(csc(θ)-1/2)=0 ?

The general solution for (cot(θ)+1)(csc(θ)-1/2)=0 is θ=(3pi)/4+pin

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(cot(θ)+1)(csc(θ)-1/2)=0

(cot (θ) + 1) (csc (θ) - \frac{1}{2}) = 0

θ = \frac{3 π}{4} + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

(cot (θ) + 1) (csc (θ) - \frac{1}{2}) = 0

Solving each part separately

cot (θ) + 1 = 0 or csc (θ) - \frac{1}{2} = 0

cot (θ) + 1 = 0 : θ = \frac{3 π}{4} + πn

cot (θ) + 1 = 0

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

csc (θ) - \frac{1}{2} = 0 :

No Solution

csc (θ) - \frac{1}{2} = 0

Move

\frac{1}{2}

to the right side

csc (θ) - \frac{1}{2} = 0

Add

\frac{1}{2}

to both sides

csc (θ) - \frac{1}{2} + \frac{1}{2} = 0 + \frac{1}{2}

Simplify

csc (θ) = \frac{1}{2}

csc (θ) = \frac{1}{2}

csc (x) \leq - 1 or csc (x) \geq 1

No Solution

Combine all the solutions

θ = \frac{3 π}{4} + πn

tan^{2} (x) = 1 + sec^{2} (x)

cos (x) = \frac{50}{86}

cos (θ) = 1.1111111

sin (5 θ) - 17 = - 18

x, z = tan (2 x - y)

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