What is the general solution for 3csc(θ)-3=0 ?

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

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3csc(θ)-3=0

3 csc (θ) - 3 = 0

θ = \frac{π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

3 csc (θ) - 3 = 0

Move

3

to the right side

3 csc (θ) - 3 = 0

Add

3

to both sides

3 csc (θ) - 3 + 3 = 0 + 3

Simplify

3 csc (θ) = 3

3 csc (θ) = 3

Divide both sides by

3

3 csc (θ) = 3

Divide both sides by

3

\frac{3 csc ( θ )}{3} = \frac{3}{3}

Simplify

csc (θ) = 1

csc (θ) = 1

General solutions for

csc (θ) = 1

csc (x)

periodicity table with

2 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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

2 sin (x) + (sin (x))^{2} = 0

2 cos (x) + 1 = 3, 0 \leq x \leq 2 π

1 - 2 sin (2 x) = 0

2 sin (x) + \frac{1}{sin ( x )} = 0

2 cos^{2} (θ) = 2 + sin (θ)

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