What is the general solution for 2sqrt(3)sec(θ)+4=0 ?

The general solution for 2sqrt(3)sec(θ)+4=0 is θ=(5pi)/6+2pin,θ=(7pi)/6+2pin

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2sqrt(3)sec(θ)+4=0

23 sec (θ) + 4 = 0

θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn

Degrees

θ = 15 0^{\circ} + 36 0^{\circ} n, θ = 21 0^{\circ} + 36 0^{\circ} n

Solution steps

23 sec (θ) + 4 = 0

Move

4

to the right side

23 sec (θ) + 4 = 0

Subtract

4

from both sides

23 sec (θ) + 4 - 4 = 0 - 4

Simplify

23 sec (θ) = - 4

23 sec (θ) = - 4

Divide both sides by

23

23 sec (θ) = - 4

Divide both sides by

23

\frac{2 3 sec ( θ )}{2 3} = \frac{- 4}{2 3}

Simplify

\frac{2 3 sec ( θ )}{2 3} = \frac{- 4}{2 3}

Simplify

\frac{2 3 sec ( θ )}{2 3} : sec (θ)

\frac{2 3 sec ( θ )}{2 3}

Divide the numbers:

\frac{2}{2} = 1

= \frac{3 sec ( θ )}{3}

Cancel the common factor:

3

= sec (θ)

Simplify

\frac{- 4}{2 3} : - \frac{2 3}{3}

\frac{- 4}{2 3}

Apply the fraction rule:

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

= - \frac{4}{2 3}

Divide the numbers:

\frac{4}{2} = 2

= - \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}

sec (θ) = - \frac{2 3}{3}

sec (θ) = - \frac{2 3}{3}

sec (θ) = - \frac{2 3}{3}

General solutions for

sec (θ) = - \frac{2 3}{3}

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn

θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn

3 sin (x) = 2

\frac{sin ( 7 4 ^{\circ} )}{8} = \frac{sin ( a )}{4}

sin (x) = sin (x - \frac{π}{2})

tan (x) = \frac{π}{3}

- 7 tan (a) - 14 = tan (a) - 6

What is the general solution for 2sqrt(3)sec(θ)+4=0 ?
The general solution for 2sqrt(3)sec(θ)+4=0 is θ=(5pi)/6+2pin,θ=(7pi)/6+2pin