What is the general solution for tan(θ/2)=(sqrt(3))/3 ?

The general solution for tan(θ/2)=(sqrt(3))/3 is θ= pi/3+2pin

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tan(θ/2)=(sqrt(3))/3

tan (\frac{θ}{2}) = \frac{3}{3}

θ = \frac{π}{3} + 2 πn

Degrees

θ = 6 0^{\circ} + 36 0^{\circ} n

Solution steps

tan (\frac{θ}{2}) = \frac{3}{3}

General solutions for

tan (\frac{θ}{2}) = \frac{3}{3}

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

\frac{θ}{2} = \frac{π}{6} + πn

\frac{θ}{2} = \frac{π}{6} + πn

Solve

\frac{θ}{2} = \frac{π}{6} + πn : θ = \frac{π}{3} + 2 πn

\frac{θ}{2} = \frac{π}{6} + πn

Multiply both sides by

2

\frac{θ}{2} = \frac{π}{6} + πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{π}{6} + 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{π}{6} + 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{π}{6} + 2 πn : \frac{π}{3} + 2 πn

2 \cdot \frac{π}{6} + 2 πn

2 \cdot \frac{π}{6} = \frac{π}{3}

2 \cdot \frac{π}{6}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{6}

Cancel the common factor:

2

= \frac{π}{3}

= \frac{π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn

x, cos (x) + 1 = 0

- 8 sin (x) = 0

sin (θ + \frac{π}{4}) = 0

2 sin^{2} (x) - 7 sin (x) + 3 = 0, 0 \leq x \leq 2 π

csc (θ) = - \frac{( 2 3 )}{3}

What is the general solution for tan(θ/2)=(sqrt(3))/3 ?
The general solution for tan(θ/2)=(sqrt(3))/3 is θ= pi/3+2pin