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

The general solution for 2/3 =sin(2θ) is θ=(0.72972…)/2+pin,θ= pi/2-(0.72972…)/2+pin

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2/3 =sin(2θ)

\frac{2}{3} = sin (2 θ)

θ = \frac{0.72972 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.72972 \dots}{2} + πn

Degrees

θ = 20.90515 \dots^{\circ} + 18 0^{\circ} n, θ = 69.09484 \dots^{\circ} + 18 0^{\circ} n

Solution steps

\frac{2}{3} = sin (2 θ)

Switch sides

sin (2 θ) = \frac{2}{3}

Apply trig inverse properties

sin (2 θ) = \frac{2}{3}

General solutions for

sin (2 θ) = \frac{2}{3}

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

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

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

Solve

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

2 θ = arcsin (\frac{2}{3}) + 2 πn

Divide both sides by

2

2 θ = arcsin (\frac{2}{3}) + 2 πn

Divide both sides by

2

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

Simplify

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

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

Solve

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

2 θ = π - arcsin (\frac{2}{3}) + 2 πn

Divide both sides by

2

2 θ = π - arcsin (\frac{2}{3}) + 2 πn

Divide both sides by

2

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

Simplify

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

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

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

Show solutions in decimal form

θ = \frac{0.72972 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.72972 \dots}{2} + πn

sin (3 x) - 2 = - 3 sin (3 x)

sin (a) = \frac{1}{5}

θ, cos (θ) = \frac{2}{3}

x, arcsin (x) = arccos (y)

sin (x) = \frac{4.9}{7.5}

What is the general solution for 2/3 =sin(2θ) ?
The general solution for 2/3 =sin(2θ) is θ=(0.72972…)/2+pin,θ= pi/2-(0.72972…)/2+pin