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

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

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sec(θ)+sqrt(2)=2sqrt(2)

sec (θ) + 2 = 22

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

sec (θ) + 2 = 22

Move

2

to the right side

sec (θ) + 2 = 22

Subtract

2

from both sides

sec (θ) + 2 - 2 = 22 - 2

Simplify

sec (θ) + 2 - 2 = 22 - 2

Simplify

sec (θ) + 2 - 2 : sec (θ)

sec (θ) + 2 - 2

Add similar elements:

2 - 2 = 0

= sec (θ)

Simplify

22 - 2 : 2

22 - 2

Add similar elements:

22 - 2 = 2

= 2

sec (θ) = 2

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

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

tan (x) sin (x) + sin (- x) = 0

- sin (A) - 1 = 3 sin (A) + 2

2 + cos (2 x) = 3 cos (x)

3 cot (θ) + 4 = 0

2 sin (3 θ) = 0

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