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

The general solution for sin(2θ)= 1/3 is θ=(0.33983…)/2+pin,θ= pi/2-(0.33983…)/2+pin

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sin(2θ)= 1/3

sin (2 θ) = \frac{1}{3}

θ = \frac{0.33983 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.33983 \dots}{2} + πn

Solution steps

sin (2 θ) = \frac{1}{3}

Apply trig inverse properties

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

Solve

2 θ = arcsin (\frac{1}{3}) + 2 πn : θ = \frac{arcsin ( \frac{1}{3} )}{2} + πn

Solve

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

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

Show solutions in decimal form

θ = \frac{0.33983 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.33983 \dots}{2} + πn

cos^{2} (θ) = 1 - sin (θ)

1 sin (3 0^{\circ}) = 1.495 sin (x)

e^{2 c o s (x) - 1} = 1

csc (θ) = - \frac{37}{12}, \frac{3 π}{2} \leq θ \leq 2 π

\frac{3}{2} = 2 \cdot 25 sin (\frac{θ}{2})

What is the general solution for sin(2θ)= 1/3 ?
The general solution for sin(2θ)= 1/3 is θ=(0.33983…)/2+pin,θ= pi/2-(0.33983…)/2+pin