What is the general solution for 0=2sin(θ) ?

The general solution for 0=2sin(θ) is θ=2pin,θ=pi+2pin

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0=2sin(θ)

0 = 2 sin (θ)

θ = 2 πn, θ = π + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

0 = 2 sin (θ)

Switch sides

2 sin (θ) = 0

Divide both sides by

2

2 sin (θ) = 0

Divide both sides by

2

\frac{2 sin ( θ )}{2} = \frac{0}{2}

Simplify

sin (θ) = 0

sin (θ) = 0

General solutions for

sin (θ) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

sin (\frac{π}{2} + x) = - cot (x)

tan (A) = - 1

sin (x) = - \frac{5}{8}

2 sin (x) = 1, 0 \leq x \leq 2 π

5 cot (x) + 4 = 9

What is the general solution for 0=2sin(θ) ?
The general solution for 0=2sin(θ) is θ=2pin,θ=pi+2pin