What is the general solution for sin(2θ)=(1.35*10)/((38)^2) ?

The general solution for sin(2θ)=(1.35*10)/((38)^2) is θ=(0.00934…)/2+pin,θ= pi/2-(0.00934…)/2+pin

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sin(2θ)=(1.35*10)/((38)^2)

sin (2 θ) = \frac{1.35 \cdot 10}{( 38 ) ^{2}}

θ = \frac{0.00934 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.00934 \dots}{2} + πn

Solution steps

sin (2 θ) = \frac{1.35 \cdot 10}{( 38 ) ^{2}}

Simplify

\frac{1.35 \cdot 10}{3 8 ^{2}} : \frac{13.5}{1444}

Apply trig inverse properties

2 θ = arcsin (\frac{13.5}{1444}) + 2 πn, 2 θ = π - arcsin (\frac{13.5}{1444}) + 2 πn

Solve

2 θ = arcsin (\frac{13.5}{1444}) + 2 πn : θ = \frac{arcsin ( \frac{13.5}{1444} )}{2} + πn

Solve

2 θ = π - arcsin (\frac{13.5}{1444}) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( \frac{13.5}{1444} )}{2} + πn

θ = \frac{arcsin ( \frac{13.5}{1444} )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( \frac{13.5}{1444} )}{2} + πn

Show solutions in decimal form

θ = \frac{0.00934 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.00934 \dots}{2} + πn

- 2 = \frac{16}{π} cos (4 \cdot x)

- (\frac{1}{2})^{2} + cos (θ^{2}) = 1

0 = 2 cos (2 θ)

2 sec^{2} (x) + tan (x) = 8

tan (θ) = \frac{48}{8}

What is the general solution for sin(2θ)=(1.35*10)/((38)^2) ?
The general solution for sin(2θ)=(1.35*10)/((38)^2) is θ=(0.00934…)/2+pin,θ= pi/2-(0.00934…)/2+pin