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

The general solution for sqrt(3)csc(θ)+2=0,0<= θ<= 2pi is θ=(4pi)/3 ,θ=(5pi)/3

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sqrt(3)csc(θ)+2=0,0<= θ<= 2pi

3 csc (θ) + 2 = 0, 0 \leq θ \leq 2 π

θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

Degrees

θ = 24 0^{\circ}, θ = 30 0^{\circ}

Solution steps

3 csc (θ) + 2 = 0, 0 \leq θ \leq 2 π

Move

2

to the right side

3 csc (θ) + 2 = 0

Subtract

2

from both sides

3 csc (θ) + 2 - 2 = 0 - 2

Simplify

3 csc (θ) = - 2

3 csc (θ) = - 2

Divide both sides by

3

3 csc (θ) = - 2

Divide both sides by

3

\frac{3 csc ( θ )}{3} = \frac{- 2}{3}

Simplify

\frac{3 csc ( θ )}{3} = \frac{- 2}{3}

Simplify

\frac{3 csc ( θ )}{3} : csc (θ)

\frac{3 csc ( θ )}{3}

Cancel the common factor:

3

= csc (θ)

Simplify

\frac{- 2}{3} : - \frac{2 3}{3}

\frac{- 2}{3}

Apply the fraction rule:

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

= - \frac{2}{3}

Rationalize

- \frac{2}{3} : - \frac{2 3}{3}

- \frac{2}{3}

Multiply by the conjugate

\frac{3}{3}

= - \frac{2 3}{3 3}

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{2 3}{3}

= - \frac{2 3}{3}

csc (θ) = - \frac{2 3}{3}

csc (θ) = - \frac{2 3}{3}

csc (θ) = - \frac{2 3}{3}

General solutions for

csc (θ) = - \frac{2 3}{3}

csc (x)

periodicity table with

2 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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

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

4 cos (t) = 3

(sin (2 x)) sin (x) = cos (x)

4 sin (x) cos (x) - 1 = 0

sin (a) = \frac{2}{5}

What is the general solution for sqrt(3)csc(θ)+2=0,0<= θ<= 2pi ?
The general solution for sqrt(3)csc(θ)+2=0,0<= θ<= 2pi is θ=(4pi)/3 ,θ=(5pi)/3