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

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

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sqrt(2)=csc(θ)

2 = csc (θ)

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

Degrees

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 36 0^{\circ} n

Solution steps

2 = csc (θ)

Switch sides

csc (θ) = 2

General solutions for

csc (θ) = 2

csc (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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

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

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

cos^{2} (15 0^{\circ}) - sin^{2} (15 0^{\circ}) = tan (x)

3 - sin (x) = 0

lo g_{8} (4) = \frac{( sin ^{2} ( x ) )}{sin ( x ) - \frac{1}{8}}

4 \cdot cos^{2} (x) - 1 = 0

tan (2 x - 5) = sin^{2} (3 0^{\circ}) + sin^{2} (6 0^{\circ})

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