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

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

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sec(θ)=-2/(sqrt(3))

sec (θ) = - \frac{2}{3}

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

Solution steps

sec (θ) = - \frac{2}{3}

Simplify

- \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}

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

tan (θ) = \frac{8}{12}

sin (t) + cos (t) = 0

0 = cos (3 x)

tan (x) = \frac{12}{14}

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

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