What is the general solution for sin(θ)= 1/5 ?

The general solution for sin(θ)= 1/5 is θ=0.20135…+2pin,θ=pi-0.20135…+2pin

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sin(θ)= 1/5

sin (θ) = \frac{1}{5}

θ = 0.20135 \dots + 2 πn, θ = π - 0.20135 \dots + 2 πn

Degrees

θ = 11.53695 \dots^{\circ} + 36 0^{\circ} n, θ = 168.46304 \dots^{\circ} + 36 0^{\circ} n

Solution steps

sin (θ) = \frac{1}{5}

Apply trig inverse properties

sin (θ) = \frac{1}{5}

General solutions for

sin (θ) = \frac{1}{5}

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

θ = arcsin (\frac{1}{5}) + 2 πn, θ = π - arcsin (\frac{1}{5}) + 2 πn

θ = arcsin (\frac{1}{5}) + 2 πn, θ = π - arcsin (\frac{1}{5}) + 2 πn

Show solutions in decimal form

θ = 0.20135 \dots + 2 πn, θ = π - 0.20135 \dots + 2 πn

sin (4 x) = sin (2 x)

cos^{2} (x) = sin^{2} (x) - \frac{2}{2}

cos (2 θ) - sin (θ) = 0

tan (θ) = - \frac{2 3}{3} sin (θ)

3 cos (x) tan (x) - cos (x) = 0

What is the general solution for sin(θ)= 1/5 ?
The general solution for sin(θ)= 1/5 is θ=0.20135…+2pin,θ=pi-0.20135…+2pin