What is the general solution for sqrt(3)tan(θ)+1=0 ?

The general solution for sqrt(3)tan(θ)+1=0 is θ=(5pi)/6+pin

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sqrt(3)tan(θ)+1=0

3 tan (θ) + 1 = 0

θ = \frac{5 π}{6} + πn

Degrees

θ = 15 0^{\circ} + 18 0^{\circ} n

Solution steps

3 tan (θ) + 1 = 0

Move

1

to the right side

3 tan (θ) + 1 = 0

Subtract

1

from both sides

3 tan (θ) + 1 - 1 = 0 - 1

Simplify

3 tan (θ) = - 1

3 tan (θ) = - 1

Divide both sides by

3

3 tan (θ) = - 1

Divide both sides by

3

\frac{3 tan ( θ )}{3} = \frac{- 1}{3}

Simplify

\frac{3 tan ( θ )}{3} = \frac{- 1}{3}

Simplify

\frac{3 tan ( θ )}{3} : tan (θ)

\frac{3 tan ( θ )}{3}

Cancel the common factor:

3

= tan (θ)

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}

tan (θ) = - \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 (x) cos (y) = \frac{1}{2} (sin (x + y) + sin (x - y))

tan (\frac{7 π}{4})

sin (- 4 5^{\circ})

sec (θ) + 2 = 0

cos (25 5^{\circ})

What is the general solution for sqrt(3)tan(θ)+1=0 ?
The general solution for sqrt(3)tan(θ)+1=0 is θ=(5pi)/6+pin