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

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

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sec(θ)+2=0

sec (θ) + 2 = 0

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

Degrees

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

sec (θ) + 2 = 0

Move

2

to the right side

sec (θ) + 2 = 0

Subtract

2

from both sides

sec (θ) + 2 - 2 = 0 - 2

Simplify

sec (θ) = - 2

sec (θ) = - 2

General solutions for

sec (θ) = - 2

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

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

cos (25 5^{\circ})

tan (13 5^{\circ})

sin (2 θ) = - \frac{3}{2}

arctan (0)

cos (\frac{7 π}{4})

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