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

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

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sqrt(3)cot(θ)+1=0,0<= θ<= 2pi

3 cot (θ) + 1 = 0, 0 \leq θ \leq 2 π

θ = \frac{2 π}{3}, θ = \frac{5 π}{3}

Degrees

θ = 12 0^{\circ}, θ = 30 0^{\circ}

Solution steps

3 cot (θ) + 1 = 0, 0 \leq θ \leq 2 π

Move

1

to the right side

3 cot (θ) + 1 = 0

Subtract

1

from both sides

3 cot (θ) + 1 - 1 = 0 - 1

Simplify

3 cot (θ) = - 1

3 cot (θ) = - 1

Divide both sides by

3

3 cot (θ) = - 1

Divide both sides by

3

\frac{3 cot ( θ )}{3} = \frac{- 1}{3}

Simplify

\frac{3 cot ( θ )}{3} = \frac{- 1}{3}

Simplify

\frac{3 cot ( θ )}{3} : cot (θ)

\frac{3 cot ( θ )}{3}

Cancel the common factor:

3

= cot (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}

cot (θ) = - \frac{3}{3}

cot (θ) = - \frac{3}{3}

cot (θ) = - \frac{3}{3}

General solutions for

cot (θ) = - \frac{3}{3}

cot (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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{2 π}{3}, θ = \frac{5 π}{3}

cos (x) = - \frac{4}{9}

7 sin (c) + 19 = 0

cos^{3} (x) = \frac{1}{2}

sin (x + 2 0^{\circ}) = cos (x - 5 0^{\circ})

2 cos^{2} (x) = sin^{2} (x)

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