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

The general solution for 1/(sqrt(3))=tan(θ) is θ= pi/6+pin

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1/(sqrt(3))=tan(θ)

\frac{1}{3} = tan (θ)

θ = \frac{π}{6} + πn

Degrees

θ = 3 0^{\circ} + 18 0^{\circ} n

Solution steps

\frac{1}{3} = tan (θ)

Switch sides

tan (θ) = \frac{1}{3}

Simplify

\frac{1}{3} : \frac{3}{3}

\frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{3}{3}

tan (θ) = \frac{3}{3}

General solutions for

tan (θ) = \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}

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

(sec^{(2)} (θ)) (cos (θ)) = 2 cos (θ)

sin (x) \cdot tan (x) + 1 = sin (x) + tan (x)

1.33 \cdot sin (5 6^{\circ}) = 1 \cdot sin (x)

cosh (x) = sinh (x)

csc (x) tan (x) = \frac{2 3}{3}

What is the general solution for 1/(sqrt(3))=tan(θ) ?
The general solution for 1/(sqrt(3))=tan(θ) is θ= pi/6+pin