What is the general solution for solvefor θ,r=sin(2θ) ?

The general solution for solvefor θ,r=sin(2θ) is θ=(arcsin(r))/2+pin,θ= pi/2+(arcsin(-r))/2+pin

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solvefor θ,r=sin(2θ)

solve for

θ, r = sin (2 θ)

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

Solution steps

r = sin (2 θ)

Switch sides

sin (2 θ) = r

Apply trig inverse properties

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

Solve

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

Solve

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

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

cos (θ) = \frac{7}{25}, \frac{3 π}{2} < θ < 2 π

5 (1 - sin (θ)) = 5

x, cos (x) = - 1

5 cos (θ) = 5 sin (- θ)

5 sin (x) + 1 = 3 sin (x)

What is the general solution for solvefor θ,r=sin(2θ) ?
The general solution for solvefor θ,r=sin(2θ) is θ=(arcsin(r))/2+pin,θ= pi/2+(arcsin(-r))/2+pin