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

The general solution for -2sin(θ/2)=0 is θ=4pin,θ=2pi+4pin

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-2sin(θ/2)=0

- 2 sin (\frac{θ}{2}) = 0

θ = 4 πn, θ = 2 π + 4 πn

Degrees

θ = 0^{\circ} + 72 0^{\circ} n, θ = 36 0^{\circ} + 72 0^{\circ} n

Solution steps

- 2 sin (\frac{θ}{2}) = 0

Divide both sides by

- 2

- 2 sin (\frac{θ}{2}) = 0

Divide both sides by

- 2

\frac{- 2 sin ( \frac{θ}{2} )}{- 2} = \frac{0}{- 2}

Simplify

sin (\frac{θ}{2}) = 0

sin (\frac{θ}{2}) = 0

General solutions for

sin (\frac{θ}{2}) = 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}

\frac{θ}{2} = 0 + 2 πn, \frac{θ}{2} = π + 2 πn

\frac{θ}{2} = 0 + 2 πn, \frac{θ}{2} = π + 2 πn

Solve

\frac{θ}{2} = 0 + 2 πn : θ = 4 πn

\frac{θ}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{θ}{2} = 2 πn

Multiply both sides by

2

\frac{θ}{2} = 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot 2 πn

Simplify

θ = 4 πn

θ = 4 πn

Solve

\frac{θ}{2} = π + 2 πn : θ = 2 π + 4 πn

\frac{θ}{2} = π + 2 πn

Multiply both sides by

2

\frac{θ}{2} = π + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 π + 2 \cdot 2 πn

Simplify

θ = 2 π + 4 πn

θ = 2 π + 4 πn

θ = 4 πn, θ = 2 π + 4 πn

cos^{3} (θ) + cos^{2} (θ) - cos (θ) - 1 = 0

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

9 sin^{2} (x) - 6 sin (x) + 1 = 0

y, ln (x) = arctan (\frac{y}{x})

cos (a) + 1 = 4 cos (a) + 1

What is the general solution for -2sin(θ/2)=0 ?
The general solution for -2sin(θ/2)=0 is θ=4pin,θ=2pi+4pin