What is the general solution for 1+sin(θ)=((sqrt(3)+2))/2 ?

The general solution for 1+sin(θ)=((sqrt(3)+2))/2 is θ=1.04719…+2pin,θ=pi-1.04719…+2pin

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1+sin(θ)=((sqrt(3)+2))/2

1 + sin (θ) = \frac{( 3 + 2 )}{2}

θ = 1.04719 \dots + 2 πn, θ = π - 1.04719 \dots + 2 πn

Degrees

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 12 0^{\circ} + 36 0^{\circ} n

Solution steps

1 + sin (θ) = \frac{( 3 + 2 )}{2}

Move

1

to the right side

1 + sin (θ) = \frac{3 + 2}{2}

Subtract

1

from both sides

1 + sin (θ) - 1 = \frac{3 + 2}{2} - 1

Simplify

sin (θ) = \frac{3 + 2}{2} - 1

sin (θ) = \frac{3 + 2}{2} - 1

Apply trig inverse properties

sin (θ) = \frac{3 + 2}{2} - 1

General solutions for

sin (θ) = \frac{3 + 2}{2} - 1

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (\frac{3 + 2}{2} - 1) + 2 πn, θ = π - arcsin (\frac{3 + 2}{2} - 1) + 2 πn

θ = arcsin (\frac{3 + 2}{2} - 1) + 2 πn, θ = π - arcsin (\frac{3 + 2}{2} - 1) + 2 πn

Show solutions in decimal form

θ = 1.04719 \dots + 2 πn, θ = π - 1.04719 \dots + 2 πn

- cos (θ) = - cos (2 θ)

cos (x) = \frac{4}{3.3 8 ^{2} + 4 ^{2}}

x, y = \frac{1}{π} arctan (\frac{x}{s}) + \frac{1}{2}

tan^{2} (θ) + 8 tan (θ) + 6 = 0

3 - sin (θ) = 5 sin (θ)

What is the general solution for 1+sin(θ)=((sqrt(3)+2))/2 ?
The general solution for 1+sin(θ)=((sqrt(3)+2))/2 is θ=1.04719…+2pin,θ=pi-1.04719…+2pin