What is the general solution for csc(θ)=(sqrt(20))/4 ?

The general solution for csc(θ)=(sqrt(20))/4 is θ=1.10714…+2pin,θ=pi-1.10714…+2pin

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csc(θ)=(sqrt(20))/4

csc (θ) = \frac{20}{4}

θ = 1.10714 \dots + 2 πn, θ = π - 1.10714 \dots + 2 πn

Radians

θ = 1.10714 \dots + 6.28318 \dots n, θ = π - 1.10714 \dots + 6.28318 \dots n

Solution steps

csc (θ) = \frac{20}{4}

Simplify

\frac{20}{4} : \frac{5}{2}

\frac{20}{4}

Factor

20 : 25

Factor

20 = 2^{2} \cdot 5

= 2^{2} \cdot 5

Apply radical rule:

= 5 2^{2}

Simplify

2^{2} : 2

2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

= 25

Factor

4 : 2^{2}

Factor

4 = 2^{2}

= \frac{2 5}{2 ^{2}}

Cancel the common factor:

2

= \frac{5}{2}

Apply trig inverse properties

csc (θ) = \frac{5}{2}

General solutions for

csc (θ) = \frac{5}{2}

csc (x) = a \Rightarrow x = arccsc (a) + 2 πn, x = π - arccsc (a) + 2 πn

θ = arccsc (\frac{5}{2}) + 2 πn, θ = π - arccsc (\frac{5}{2}) + 2 πn

θ = arccsc (\frac{5}{2}) + 2 πn, θ = π - arccsc (\frac{5}{2}) + 2 πn

Show solutions in decimal form

θ = 1.10714 \dots + 2 πn, θ = π - 1.10714 \dots + 2 πn

sin (x + 1 0^{\circ}) = cos (x + 2 0^{\circ})

c^{2} = b^{2} + a^{2} - 2 (b) (a) cos (θ)

5 cos (θ) = 2 + cos (θ)

cos (x + π) = 1

x, sin (4 x) = - 2 (sin (2 x))

What is the general solution for csc(θ)=(sqrt(20))/4 ?
The general solution for csc(θ)=(sqrt(20))/4 is θ=1.10714…+2pin,θ=pi-1.10714…+2pin