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

The general solution for tan(θ)= 1/(-sqrt(3)) is θ=(5pi)/6+pin

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tan(θ)= 1/(-sqrt(3))

tan (θ) = \frac{1}{- 3}

θ = \frac{5 π}{6} + πn

Solution steps

tan (θ) = \frac{1}{- 3}

Simplify

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

\frac{1}{- 3}

Apply the fraction rule:

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

= - \frac{1}{3}

Rationalize

- \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}

= - \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{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

sin^{2} (θ) - cos^{2} (θ) = sin (θ)

4 sin (c) + 3 = sin (c) + 1

(tan (x) + 1) (3 tan (x) - 1) = 0

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

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

What is the general solution for tan(θ)= 1/(-sqrt(3)) ?
The general solution for tan(θ)= 1/(-sqrt(3)) is θ=(5pi)/6+pin