What is the general solution for sqrt(3)tan(θ)+1=0,0<= θ<= 2pi ?

The general solution for sqrt(3)tan(θ)+1=0,0<= θ<= 2pi is θ=(5pi)/6 ,θ=(11pi)/6

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sqrt(3)tan(θ)+1=0,0<= θ<= 2pi

3 tan (θ) + 1 = 0, 0 \leq θ \leq 2 π

θ = \frac{5 π}{6}, θ = \frac{11 π}{6}

Degrees

θ = 15 0^{\circ}, θ = 33 0^{\circ}

Solution steps

3 tan (θ) + 1 = 0, 0 \leq θ \leq 2 π

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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{5 π}{6}, θ = \frac{11 π}{6}

tan (x) = tan^{2} (x)

y, sin (x y) = x

tan (x) sec (x) = tan (x)

1 - sin (2 x) = cos (2 x)

sin (2 x - 5 \frac{π}{6}) = - \frac{1}{2}, 0 \leq x \leq 2 π

What is the general solution for sqrt(3)tan(θ)+1=0,0<= θ<= 2pi ?
The general solution for sqrt(3)tan(θ)+1=0,0<= θ<= 2pi is θ=(5pi)/6 ,θ=(11pi)/6