What is the general solution for sin(θ)=0.321 ?

The general solution for sin(θ)=0.321 is θ=0.32678…+2pin,θ=pi-0.32678…+2pin

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sin(θ)=0.321

sin (θ) = 0.321

θ = 0.32678 \dots + 2 πn, θ = π - 0.32678 \dots + 2 πn

Solution steps

sin (θ) = 0.321

Apply trig inverse properties

θ = arcsin (0.321) + 2 πn, θ = π - arcsin (0.321) + 2 πn

Show solutions in decimal form

θ = 0.32678 \dots + 2 πn, θ = π - 0.32678 \dots + 2 πn

sin (x + 7 5^{\circ}) = \frac{3}{2}

tan (\frac{x}{6}) + 3 = 0

A, (1 + tan (A)) (1 + tan (B)) = 2

5 cos (2 t) cos (t) = 0

k, 6 (- cos (\frac{k}{2}) + 1) = 1.5

What is the general solution for sin(θ)=0.321 ?
The general solution for sin(θ)=0.321 is θ=0.32678…+2pin,θ=pi-0.32678…+2pin