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

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

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sqrt(2)=sec(θ)

2 = sec (θ)

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Degrees

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

2 = sec (θ)

Switch sides

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{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

sin (x) = \frac{1}{8}, sin (2 x)

x, csc (x) = 2

\frac{3}{2} sec (x) = 3

9 (cos (a))^{2} - 9 (sin (a))^{2} = 0

1 - 2 cos^{2} (x) = 0, 0 \leq x \leq 2 π

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