What is the general solution for (sqrt(3))/3 =-1/3*tan(θ) ?

The general solution for (sqrt(3))/3 =-1/3*tan(θ) is θ=(2pi)/3+pin

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(sqrt(3))/3 =-1/3*tan(θ)

\frac{3}{3} = - \frac{1}{3} \cdot tan (θ)

θ = \frac{2 π}{3} + πn

Degrees

θ = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

\frac{3}{3} = - \frac{1}{3} tan (θ)

Switch sides

- \frac{1}{3} tan (θ) = \frac{3}{3}

Multiply both sides by

- 3

- \frac{1}{3} tan (θ) = \frac{3}{3}

Multiply both sides by

- 3

(- \frac{1}{3} tan (θ)) (- 3) = \frac{3 ( - 3 )}{3}

Simplify

(- \frac{1}{3} tan (θ)) (- 3) = \frac{3 ( - 3 )}{3}

Simplify

(- \frac{1}{3} tan (θ)) (- 3) : tan (θ)

(- \frac{1}{3} tan (θ)) (- 3)

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1}{3} tan (θ) \cdot 3

Multiply fractions:

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

= \frac{1 \cdot 3}{3} tan (θ)

Cancel the common factor:

3

= tan (θ) \cdot 1

Multiply:

tan (θ) \cdot 1 = tan (θ)

= tan (θ)

Simplify

\frac{3 ( - 3 )}{3} : - 3

\frac{3 ( - 3 )}{3}

Remove parentheses:

(- a) = - a

= \frac{- 3 \cdot 3}{3}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{3 \cdot 3}{3}

Divide the numbers:

\frac{3}{3} = 1

= - 3

tan (θ) = - 3

tan (θ) = - 3

tan (θ) = - 3

General solutions for

tan (θ) = - 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}

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

tan (x + \frac{π}{3}) = 3

16 sin^{2} (x) = 1

x, - 1 \cdot \frac{cos ( x )}{sin ( x )} = 0

8 cos (θ) - 4 = 0

2 (1 + cos (θ)) = 0

What is the general solution for (sqrt(3))/3 =-1/3*tan(θ) ?
The general solution for (sqrt(3))/3 =-1/3*tan(θ) is θ=(2pi)/3+pin