What is the general solution to dydθ=sin(θ)\frac{dy}{d\theta} = \sin(\theta)dθdy=sin(θ)?
y=cos(θ)+Cy = \cos(\theta) + Cy=cos(θ)+C
y=−cos(θ)+Cy = -\cos(\theta) + Cy=−cos(θ)+C
y=sin(θ)+Cy = \sin(\theta) + Cy=sin(θ)+C
y=−sin(θ)+Cy = -\sin(\theta) + Cy=−sin(θ)+C