What is the general solution for (-2sqrt(3))/3 =sec(θ) ?

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

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(-2sqrt(3))/3 =sec(θ)

\frac{- 2 3}{3} = sec (θ)

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

\frac{- 2 3}{3} = sec (θ)

Switch sides

sec (θ) = \frac{- 2 3}{3}

Apply the fraction rule:

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

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

\frac{- 2 sin ( x ) - 1}{( 2 + sin ( x ) ) ^{2}} = 0

cot (x) = - 8

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

- 1 + sin^{2} (a) + cos^{2} (a) = cos^{2} (a)

\frac{sin ( x )}{1 + cos ( x )} = 3 cot (\frac{x}{2})

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