What is the general solution for sqrt(2)csc(θ)-2=0 ?

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

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sqrt(2)csc(θ)-2=0

2 csc (θ) - 2 = 0

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) - 2 = 0

Move

2

to the right side

2 csc (θ) - 2 = 0

Add

2

to both sides

2 csc (θ) - 2 + 2 = 0 + 2

Simplify

2 csc (θ) = 2

2 csc (θ) = 2

Divide both sides by

2

2 csc (θ) = 2

Divide both sides by

2

\frac{2 csc ( θ )}{2} = \frac{2}{2}

Simplify

\frac{2 csc ( θ )}{2} = \frac{2}{2}

Simplify

\frac{2 csc ( θ )}{2} : csc (θ)

\frac{2 csc ( θ )}{2}

Cancel the common factor:

2

= csc (θ)

Simplify

\frac{2}{2} : 2

\frac{2}{2}

Apply radical rule:

2 = 2^{\frac{1}{2}}

= \frac{2}{2 ^{\frac{1}{2}}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{2 ^{1}}{2 ^{\frac{1}{2}}} = 2^{1 - \frac{1}{2}}

= 2^{1 - \frac{1}{2}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= 2^{\frac{1}{2}}

Apply radical rule:

2^{\frac{1}{2}} = 2

= 2

csc (θ) = 2

csc (θ) = 2

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

sin (θ) = \frac{5}{4}

100 = 5040 cos (θ)

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

cos (x) (csc (x) + 2) = 0

arctan (θ) = \frac{π}{4}

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