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

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

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tan(θ)+1/(sqrt(3))=0

tan (θ) + \frac{1}{3} = 0

θ = \frac{5 π}{6} + πn

Degrees

θ = 15 0^{\circ} + 18 0^{\circ} n

Solution steps

tan (θ) + \frac{1}{3} = 0

Move

\frac{1}{3}

to the right side

tan (θ) + \frac{1}{3} = 0

Subtract

\frac{1}{3}

from both sides

tan (θ) + \frac{1}{3} - \frac{1}{3} = 0 - \frac{1}{3}

Simplify

tan (θ) + \frac{1}{3} - \frac{1}{3} = 0 - \frac{1}{3}

Simplify

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

tan (θ) + \frac{1}{3} - \frac{1}{3}

Add similar elements:

\frac{1}{3} - \frac{1}{3} = 0

= tan (θ)

Simplify

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

0 - \frac{1}{3}

0 - \frac{1}{3} = - \frac{1}{3}

= - \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

12 cos^{2} (x) - 3 = 0

csc^{2} (θ) + 4 csc (θ) - 5 = 0

cos^{2} (x) - cos (x) = 0, 0 \leq x \leq 2 π

\frac{sin ( 2 x )}{cos ( x )} = 2, 0 \leq x \leq π

sec (θ) = - 4

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